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Algebra

Quadratic Function Equation

Contents

Introduction

The quadratic function equations are such polynomial equations that can be either of one variable and/or more than one variable, where the maximum value of the exponent of any variable is not more than two. The highest value of the exponent is two therefore the equation is called the quadratic equation. In other words, among the all variables in the equation if any of them has a maximum exponent value equal to two, then that equation is called the quadratic equation. A quadratic equation is a form of the algebraic equation, that is intrinsically defined in the form of {\rm{f}}\left( {\rm{x}} \right){\rm{ = }}{{\rm{x}}^{\rm{2}}}, and therefore, this function maps the points which can be defined with the \left( {{\rm{x, }}{{\rm{x}}^{\rm{2}}}} \right) coordinates. 

The quadratic function equation is important because of its practical applications in daily life. In the launch of a rocket, its trajectory can be predicted by the simple quadratic function equation, isn’t it fascinating? The other uses of the quadratic function involve the calculation of speed, the area under the curve, an estimate of the projectile motion of any object, and the loss and profit in the area of economics. 

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The standard form of the quadratic function equation

The standard form of the quadratic function equation can be expressed as,

{\rm{f}}\left( {\rm{x}} \right){\rm{ = a}}{{\rm{x}}^{\rm{2}}}{\rm{ + bx + c}} 

where {\rm{a,}}\;{\rm{b,}}\;{\rm{and}}\;{\rm{c}} can have the values of any real number except

    \[{\rm{a}} \ne 0\]

, and x is the variable of the equation.

The standard form of the equation is used to generalize the outcome of the given quadratic function equations. Some examples of quadratic function equations are,

\begin{array}{l}{\rm{f}}\left( {\rm{x}} \right){\rm{ = 3}}{{\rm{x}}^{\rm{2}}}{\rm{,}}\;{\rm{where}}\;{\rm{a = 3,}}\;{\rm{b = 0, and c = 0}}\\{\rm{f}}\left( {\rm{x}} \right){\rm{ = 3}}{{\rm{x}}^{\rm{2}}}{\rm{ - 4x,}}\;{\rm{where}}\;{\rm{a = 3,}}\;{\rm{b = - 4, and c = 0}}\\{\rm{f}}\left( {\rm{x}} \right){\rm{ = 3}}{{\rm{x}}^{\rm{2}}}{\rm{ - 4x + 5,}}\;{\rm{where}}\;{\rm{a = 3,}}\;{\rm{b = - 4, and c = 5}}\end{array}

The role of the constants {\rm{a,}}\;{\rm{b,}}\;{\rm{and}}\;{\rm{c}}in shaping the graph of the quadratic function

Since the quadratic function equations have at least one of their variables with the power of the exponent equal to 2 therefore the quadratic function equations are given in the shape of the parabola, which is always symmetric about a given axis of symmetry and has one vertex. The vertex and axis of symmetry of the parabolic curve are defined in the below graph. Therefore, we can define the shape of the quadratic function equations using the standard form of the quadratic function equations.

Figure: The vertex and axis of symmetry of the parabola

For the case, when variables{\rm{b,}}\;{\rm{and}}\;{\rm{c}}are equal to zero. Since the variable {\rm{a}} can not be zero, therefore, we can make the graphs of the function,

{\rm{f}}\left( {\rm{x}} \right){\rm{ = a}}{{\rm{x}}^{\rm{2}}}

such that, if the coefficient

    \[{\rm{a}}\]

is positive, then it gives an upward curve (towards the positive axis) while if it is negative then it gives a downward curve (towards the negative axis). In the below graphs the quadratic function equation {\rm{f}}\left( {\rm{x}} \right){\rm{ = 2}}{{\rm{x}}^{\rm{2}}},\;{\rm{a}} = 2, gives an upward curve (solid red curve), while the equation {\rm{f}}\left( {\rm{x}} \right){\rm{ = - 2}}{{\rm{x}}^{\rm{2}}},\;{\rm{a}} = - 2 gives a downward curve (solid blue curve), therefore the sign of the coefficient {\rm{a}}in the quadratic function equation determines the direction of the opening of the parabola. 

Figure 1: Graph of the functions {\rm{f}}\left( {\rm{x}} \right){\rm{ = 2}}{{\rm{x}}^{\rm{2}}}\;{\rm{and}}\;{\rm{f}}\left( {\rm{x}} \right){\rm{ = - 2}}{{\rm{x}}^{\rm{2}}}, in the red and blue solid curves, respectively.

Further, if we see the graphs, where we have added another term such that the quadratic function equations become in the form of,

{\rm{f}}\left( {\rm{x}} \right){\rm{ = a}}{{\rm{x}}^{\rm{2}}} + {\rm{bx}}

Therefore, if we see the below graphs in Fig. 2 of the quadratic function equations of {\rm{f}}\left( {\rm{x}} \right){\rm{ = 2}}{{\rm{x}}^{\rm{2}}} + 5{\rm{x, a = 2,}}\;{\rm{and}}\;{\rm{b = 5}}(red solid curve) and {\rm{f}}\left( {\rm{x}} \right){\rm{ = - 2}}{{\rm{x}}^{\rm{2}}} + 5{\rm{x, a = 2,}}\;{\rm{and}}\;{\rm{b = 5}}(blue solid curve), and if we compare this graph with the graphs in Fig. 1, we can see that the positive value of the coefficient {\rm{b}}shifts the curves in the negative x-axis and negative f(x) -axis if the coefficient {\rm{a}}is positive, while shifts to the positive x-axis and positive f(x)-axis if the coefficient {\rm{a}}is negative.

Figure 2: Graph of the functions {\rm{f}}\left( {\rm{x}} \right){\rm{ = 2}}{{\rm{x}}^{\rm{2}}} + 5{\rm{x}}\;{\rm{and}}\;{\rm{f}}\left( {\rm{x}} \right){\rm{ = - 2}}{{\rm{x}}^{\rm{2}}} + 5{\rm{x}}, in the red and blue solid curves, respectively.

Similarly, if we see the below graphs in Fig. 3 of the quadratic function equations of {\rm{f}}\left( {\rm{x}} \right){\rm{ = 2}}{{\rm{x}}^{\rm{2}}} - 5{\rm{x, a = 2,}}\;{\rm{and}}\;{\rm{b = - 5}}(red solid curve) and {\rm{f}}\left( {\rm{x}} \right){\rm{ = - 2}}{{\rm{x}}^{\rm{2}}} - 5{\rm{x, a = 2,}}\;{\rm{and}}\;{\rm{b = - 5}}(blue solid curve), and if we compare this graph with the graphs in Fig. 1, we can see that the negative value of the coefficient {\rm{b}}shifts the curves in the positive x-axis and negative f(x)-axis if the coefficient {\rm{a}}is positive, while shifts to the negative x-axis and positive f(x)-axis if the coefficient {\rm{a}}is negative. 

Thus the presence of the coefficient {\rm{b}}in the quadratic function shifts the curve vertically as well as horizontally and therefore shifts the axis of symmetry as well as the vertex of the parabolic curves.

Figure 2: Graph of the functions {\rm{f}}\left( {\rm{x}} \right){\rm{ = 2}}{{\rm{x}}^{\rm{2}}} - 5{\rm{x}}\;{\rm{and}}\;{\rm{f}}\left( {\rm{x}} \right){\rm{ = - 2}}{{\rm{x}}^{\rm{2}}} - 5{\rm{x}}, in the red and blue solid curves, respectively.

Further, if we see the below graphs in Fig. 4 of the quadratic function equations of {\rm{f}}\left( {\rm{x}} \right){\rm{ = 2}}{{\rm{x}}^{\rm{2}}} - 5{\rm{x + 4, a = 2,}}\;{\rm{b = - 5,}}\;{\rm{and c = 4}}(red solid curve) and {\rm{f}}\left( {\rm{x}} \right){\rm{ = - 2}}{{\rm{x}}^{\rm{2}}} - 5{\rm{x + 4, a = - 2,}}\;{\rm{b = - 5,}}\;{\rm{and c = 4}}(blue solid curve), and if we compare this graph with the graphs in Fig. 3, we can see that the positive value of the coefficient {\rm{c}}, shifts the curves in the positive f(x)-axis, corresponding to the value of the coefficient {\rm{c}}. Thus the presence of the coefficient {\rm{c}}in the quadratic function shifts the curve vertically on the f(x)-axis similar to the sign of the coefficient {\rm{c}}.

Figure 4: Graph of the functions {\rm{f}}\left( {\rm{x}} \right){\rm{ = 2}}{{\rm{x}}^{\rm{2}}} - 5{\rm{x + 4}}\;{\rm{and}}\;{\rm{f}}\left( {\rm{x}} \right){\rm{ = - 2}}{{\rm{x}}^{\rm{2}}} - 5{\rm{x + 4}}, in the red and blue solid curves, respectively.

How to use the quadratic function equation to solve problems

For the solution of the given quadratic function equation of the standard form {\rm{f}}\left( {\rm{x}} \right){\rm{ = a}}{{\rm{x}}^{\rm{2}}}{\rm{ + bx + c}}, first, we need to calculate the value of the equation {{\rm{b}}^{\rm{2}}}{\rm{ - 4ac}}, which is called the discriminant of the quadratic function equation. 

If, \left( {{{\rm{b}}^{\rm{2}}}{\rm{ - 4ac}}} \right) = 0, then we get two equal solutions of the quadratic function equation.

If, \left( {{{\rm{b}}^{\rm{2}}}{\rm{ - 4ac}}} \right) > 1, then we get two distinct real solutions of the quadratic function equation.

If, \left( {{{\rm{b}}^{\rm{2}}}{\rm{ - 4ac}}} \right) < 0, then we get two complex solutions of the quadratic function equation.

Step-by-step instructions for calculating the value of{\rm{y}} a given value of{\rm{x}}

Step 1: Substitute the given value of x in the given quadratic equation {\rm{y = a}}{{\rm{x}}^{\rm{2}}}{\rm{ + bx + c}}

Step 2: Calculate the value on the right-hand side of the given equation, this obtained value is the value of y for the given value of x.

Step-by-step instructions for calculating the roots of the given quadratic equation

Step 1: Determine the coefficients {\rm{a,}}\;{\rm{b,}}\;{\rm{and}}\;{\rm{c}} of the given quadratic function equation by comparing it with the standard form of the quadratic function equation {\rm{f}}\left( {\rm{x}} \right){\rm{ = a}}{{\rm{x}}^{\rm{2}}}{\rm{ + bx + c}}.

Step 2: Calculate the value of {{\rm{b}}^{\rm{2}}}{\rm{ - 4ac}}, to estimate the nature of the solution of the given quadratic function equation.

Step 3: Use the quadratic formula to obtain the solutions of the given quadratic function equation, such that,

{\rm{x = }}\frac{{{\rm{ - b}} \pm \sqrt {{{\rm{b}}^{\rm{2}}}{\rm{ - 4ac}}} }}{{2{\rm{a}}}}

This formula will give the solutions of the given quadratic function equation.

Example: Finding the value of{\rm{y}}when{\rm{x = 1}}for the equation {\rm{y = 2}}{{\rm{x}}^{\rm{2}}}{\rm{ + 3x - 4}}.

Solution: 

Step 1: Substitute the value of {\rm{x = 1}}, in the above given quadratic function equation, {\rm{y = 2}}{{\rm{x}}^{\rm{2}}}{\rm{ + 3x - 4}}

Step 2: Calculate the value of the equation on the right-hand side of the equation {\rm{2}}{{\rm{x}}^{\rm{2}}}{\rm{ + 3x - 4 = 2}}{\left( 1 \right)^2} + 3\left( 1 \right) - 4 = 2 + 3 - 4 = 5 - 4 = 1

Thus, the given quadratic function equation has a value of {\rm{y = 1}} for the given value of {\rm{x = 1}}.

Graphing the quadratic function equation

Since the graph of a quadratic function equation is a parabola, then we can interpret the graph of the equation using certain steps as follows,

Step 1: Determine the vertex of the parabola. 

The vertex of the parabola representing the quadratic function equation can be determined by the coordinates \left( {{\rm{x,f}}\left( {\rm{x}} \right)} \right), such that, for the given values of coefficients {\rm{a,}}\;{\rm{b,}}\;{\rm{and}}\;{\rm{c}}, we get the coordinates

    \[\left( {\frac{{ - {\rm{b}}}}{{{\rm{2a}}}},{\rm{f}}\left( {\frac{{ - {\rm{b}}}}{{{\rm{2a}}}}} \right)} \right)\]

.

Step 2: Make a table between the variable x and the value of the function f(x), where you determine the value of f(x) at least for the five given values of x. Note that, determine the values of x in such a way that at least two values of x lie on the left and right side of the vertex.

Step 3: After obtaining these coordinates, plot them on the coordinate plane using the graph and then join these points with a curve and that gives the perfect shape of the curve for the given quadratic function equation.

Using the graph of a quadratic function equation to understand its properties, such as the y-intercept, the x-intercepts (if any), and the vertex

The standard form of the quadratic function equation can be used to estimate the x- and y-intercepts as well as the vertex of the quadratic function equation.

Determination of the x-intercept:

The axis of symmetry of the quadratic function equation is known as the x-intercept of the quadratic function equation, and can be expressed as,

{\rm{x = }}\frac{{ - {\rm{b}}}}{{{\rm{2a}}}}

Determination of the y-intercept:

For the determination of the y-intercept, we substitute the value of x=0 in the given quadratic function equation and therefore the value of the coefficient c determines the value of the y-intercept, and given as,

{\rm{y = f}}\left( {\rm{x}} \right){\rm{ = c}}

Vertex of the quadratic function equation:

The vertex of the parabola representing the quadratic function equation can be determined by the coordinates \left( {{\rm{x,f}}\left( {\rm{x}} \right)} \right), such that, for the given values of coefficients {\rm{a,}}\;{\rm{b,}}\;{\rm{and}}\;{\rm{c}}, we get the coordinates

    \[\left( {\frac{{ - {\rm{b}}}}{{{\rm{2a}}}},{\rm{f}}\left( {\frac{{ - {\rm{b}}}}{{{\rm{2a}}}}} \right)} \right)\]

.

Applications of the quadratic function equation

The quadratic function equations can be used to estimate the height of the projectile motion because the equation of the projectile motion is also a quadratic function equation, expressed as,

{\rm{y = x}}\left( {{\rm{tan\theta }}} \right){\rm{ - }}\left( {\frac{{\rm{g}}}{{{{\rm{u}}^{\rm{2}}}{\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{\theta }}}}} \right){{\rm{x}}^{\rm{2}}}

This equation can be used to estimate the maximum height of an object under the projectile motion.

The quadratic function equations are also used for the production and cost estimate and are named as quadratic cost functions. These functions are also given as,

{\rm{Total Cost = a + bQ + c}}{{\rm{Q}}^{\rm{2}}}

Where a, b, and c are the real constants and Q is the variable that shows the quantity of the goods produced.

Conclusion

The quadratic function equations are the equations in which at least one of the variables have the maximum value of the exponent equal to two. The quadratic function equations have uses in the prediction of the trajectory of the object under the projectile motion, in space exploration, and areas of economics. The quadratic function equations have two roots or solutions that can be determined using the quadratic formula. Moreover, the quadratic function equations present a parabolic curve, where the a, b, and c constants are the key deciding factors of its shape and location.

  • The standard form of the quadratic function equation is given by the {\rm{f(x) = a}}{{\rm{x}}^{\rm{2}}}{\rm{ + bx + c}}, where a, b and c are the real numbers and a can not have the zero value.
  • The quadratic function equations have the shape of a parabola because they have a square variable term.
  • The determinant of the quadratic function equation is given as the {{\rm{b}}^{\rm{2}}}{\rm{ - 4ac}}, which is utilized to determine the nature of the roots of the given quadratic equation.
  • The roots of the given quadratic function equation can be determined by the formula {\rm{x = }}\frac{{{\rm{ - b}} \pm \sqrt {{{\rm{b}}^{\rm{2}}}{\rm{ - 4ac}}} }}{{2{\rm{a}}}}.

Solved Examples

Q.1. Find the roots of the equation {\rm{f(x) = 2}}{{\rm{x}}^{\rm{2}}}{\rm{ + 3x - 4}}.

Solution:

Step 1: After comparing the given equation with the standard form, we get, 

{\rm{a = 2,}}\;{\rm{b = 3,}}\;{\rm{and}}\;{\rm{c = - 4}}.

Step 2: Calculation of {{\rm{b}}^{\rm{2}}}{\rm{ - 4ac}} gives a value of 41, which is greater than 0, and therefore this equation will have two distinct real solutions.

Step 3: Using the quadratic formula, we get, 

{\rm{x = }}\frac{{{\rm{ - 3}} \pm \sqrt {{3^{\rm{2}}}{\rm{ - 4}} \times {\rm{2}} \times \left( { - 4} \right)} }}{{2 \times 2}} = \frac{{{\rm{ - 3}} \pm \sqrt {{\rm{9 + }}32} }}{4} = \frac{{{\rm{ - 3}} \pm \sqrt {41} }}{4}

Thus the two real solutions of the given quadratic function equation are, 

\frac{{{\rm{ - 3 + }}\sqrt {41} }}{4}\;{\rm{and}}\;\frac{{{\rm{ - 3 - }}\sqrt {41} }}{4}.

Q.2. Find the vertex of the given quadratic equation{\rm{f(x) = 5}}{{\rm{x}}^{\rm{2}}}{\rm{ + 4x - 6}}.

Solution:

Step 1: Comparing the given equation with the standard form, we get the value of the coefficient {\rm{a = 5,}}\;{\rm{b = 4,}}\;{\rm{and}}\;{\rm{c = - 6}}. Therefore, the x-coordinate of the vertex is given as, 

{\rm{x = }}\frac{{ - {\rm{b}}}}{{{\rm{2a}}}} = \frac{{ - 4}}{{2 \times 10}} = - \frac{1}{5}

For the given value of x, the value of the y-coordinate will be,

\begin{array}{l}{\rm{f}}\left( { - \frac{1}{5}} \right) = 5{\left( {\frac{{ - 1}}{5}} \right)^2} + 4\left( { - \frac{1}{5}} \right) - 6 = \frac{5}{{25}} - \frac{4}{5} - 6\\{\rm{f}}\left( { - \frac{1}{5}} \right) = \frac{1}{5} - \frac{4}{5} - 6 = \frac{{ - 3}}{5} - 6 = \frac{{ - 33}}{5}\end{array}

Therefore, the coordinates of the vertex are \left( { - \frac{1}{5}, - \frac{{33}}{5}} \right)

Q.3. Find the x and y-intercepts of the given quadratic equation{\rm{f(x) = 2}}{{\rm{x}}^{\rm{2}}}{\rm{ + 4x + 3}}.

Solution: 

After comparing the given equation with the standard form, we get, 

{\rm{a = 2,}}\;{\rm{b = 4,}}\;{\rm{and}}\;{\rm{c = 3}}.

The x-intercept of the equation is given by the axis of symmetry and can be expressed as,

\begin{array}{l}{\rm{x = }}\frac{{ - {\rm{b}}}}{{2{\rm{a}}}} = \frac{{ - 4}}{4}\\{\rm{x}} = - 1\end{array}

The y-intercept of the given quadratic function equation is given as,

\begin{array}{l}{\rm{y}} = {\rm{c}}\\{\rm{y = 3}}\end{array}

Thus, the x- and y- intercepts can be given as, x=-1,  and y=3.

Q.4. Find the roots of the given quadratic function equation {\rm{f(x) = 4}}{{\rm{x}}^{\rm{2}}}{\rm{ + 6x + 2}}.

Solution:

Step 1: After comparing the given equation with the standard form, we get, 

{\rm{a = 4,}}\;{\rm{b = 6,}}\;{\rm{and}}\;{\rm{c = 2}}.

Step 2: Calculation of {{\rm{b}}^{\rm{2}}}{\rm{ - 4ac}} gives a value of 20, which is greater than 0, and therefore this equation will have two distinct real solutions.

Step 3: Using the quadratic formula, we get, 

\begin{array}{l}{\rm{x = }}\frac{{{\rm{ - 6}} \pm \sqrt {{6^{\rm{2}}}{\rm{ - 4}} \times 2 \times \left( 4 \right)} }}{{2 \times 4}} = \frac{{{\rm{ - 6}} \pm \sqrt {{\rm{36 - }}32} }}{8} = \frac{{{\rm{ - 6}} \pm \sqrt 4 }}{4}\\{\rm{x}} = \frac{{{\rm{ - 6}} \pm 2}}{4}\end{array}

Thus the two real solutions of the given quadratic function equation are, 

- 1\;{\rm{and}}\; - 2.

Q.5. Find the roots of the given quadratic function equation {\rm{f(x) = }}{{\rm{x}}^{\rm{2}}}{\rm{ + 2x - 4}}.

Solution:

Step 1: After comparing the given equation with the standard form, we get, 

{\rm{a = 1,}}\;{\rm{b = 2,}}\;{\rm{and}}\;{\rm{c = - 4}}.

Step 2: Calculation of {{\rm{b}}^{\rm{2}}}{\rm{ - 4ac}} gives a value of 20, which is greater than 0, and therefore this equation will have two distinct real solutions.

Step 3: Using the quadratic formula, we get, 

\begin{array}{l}{\rm{x = }}\frac{{{\rm{ - 2}} \pm \sqrt {{2^{\rm{2}}}{\rm{ - 4}} \times 1 \times \left( { - 4} \right)} }}{{2 \times 1}} = \frac{{{\rm{ - 2}} \pm \sqrt {{\rm{4 + 16}}} }}{2} = \frac{{{\rm{ - 2}} \pm \sqrt {20} }}{2} = \frac{{{\rm{ - 2}} \pm 2\sqrt 5 }}{2}\\{\rm{x}} = - 1 \pm \sqrt 5 \end{array}

Thus the two real solutions of the given quadratic function equation are, 

- 1 + \sqrt 5 \;{\rm{and}}\; - 1 - \sqrt 5.

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FAQs

1. What is the most basic quadratic function equation?

Ans: The most basic quadratic function equation is expressed as, {\rm{f}}\left( {\rm{x}} \right){\rm{ = }}{{\rm{x}}^{\rm{2}}}. Comparing to the standard form the values of coefficients {\rm{a,}}\;{\rm{b,}}\;{\rm{and}}\;{\rm{c}} are {\rm{1,}}\;0{\rm{,}}\;{\rm{and}}\;0, respectively.

2. What is the condition that results in the non-real solutions of the quadratic function equation?

Ans: For the given quadratic function equation, if the value of the determinant {{\rm{b}}^{\rm{2}}}{\rm{ - 4ac}} is less than zero, then the quadratic function equation will have non-real or complex solutions.

3. What is the difference between a polynomial and a quadratic equation?

Ans: The quadratic equation is a subset of the polynomial equations, since quadratic equations have the maximum value of the exponents of the given variable as two, while for a polynomial equation, the exponent of the variable can have any real values.

4. What are the single-variable and multivariable quadratic equations?

Ans: In the single-variable quadratic equation, there is only one variable, e.g. x or y, and the square term in that variable. While in the case of the multivariable quadratic function equation, there can be more than one variable, e.g., x and y, and the value of the exponent two can be either of them or both of them.

5. What is the shape of the standard form of the given equation?

Ans: A given quadratic equation always has the shape of a parabola, because of the presence of the square term in the equation.

References

  1. (n.d.). Quadratic Functions. Retrieved January 13, 2023, from http://dl.uncw.edu/digilib/Mathematics/Algebra/mat111hb/PandR/quadratic/quadratic.html
  2. Quadratic Equation — from Wolfram MathWorld. (n.d.). Wolfram MathWorld. Retrieved January 13, 2023, from https://mathworld.wolfram.com/QuadraticEquation.html
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