Graphing Quadratic Equations

The solutions of a quadratic equation are the

x

-intercepts of the graph of the related function.

A quadratic equation is a polynomial equation, where

a

b

, and

c

are real numbers and

a\neq0

, that can be written in the form:

ax^2+bx+c=0

A solution of an equation is a value of the variable that makes the equation true, which means both sides are equal. A quadratic equation can have zero, one, or two real solutions. A solution to a quadratic equation is also called a root. The real solutions of a quadratic equation represent zeros of the related quadratic function:

f(x)=ax^2+bx+c

A zero of a function is any input value that makes the output of the function equal zero. The zeros also describe the

x

-intercepts of the graph of the function. An x-intercept is the value of

x

, where the graph touches or crosses the

x

-axis.

To solve a quadratic equation by graphing, graph the related function $f(x)=ax^2+bx+c$ . The real solutions are represented by the $x$ -intercepts of the graph. A quadratic function may have zero, one, or two $x$ -intercepts. For example:

For the equation $x^2+x-6=0$ , the graph of its related function, $f(x)=x^2+x-6$ , has two $x$ -intercepts, $x=-3$ and $x=2$ . So, the equation has two real solutions: $x=-3$ and $x=2$ .
For the equation $x^2-8x+16=0$ , the graph of its related function, $g(x)=x^2-8x+16$ , has one $x$ -intercept at $x=4$ . So, the equation has one real solution at $x=4$ .
For the equation $-x^2-12x-38=0$ , the graph of the related function, $h(x)=-x^2-12x-38$ , has no $x$ -intercepts. So, the equation has no real solutions.

The

x

-intercepts of a graph represent real solutions of the related equation.

On a graphing calculator, TRACE can be used to estimate the

x

-intercepts. When graphing by hand, it may be difficult to identify exact solutions. Even when the graph appears to pass through a point, verify that the

x

-intercepts read from the graph are solutions of the equation.

Solutions or Roots	Zeros	$x$ -Intercepts
An equation has solutions, which are also called roots. A solution is a value of the variable that makes the equation true.	A function has zeros. A zero is an input value that makes the value of the function zero.	A graph has $x$ -intercepts. An $x$ -intercept is an $x$ -coordinate of a point where the graph touches or crosses the $x$ -axis.

Graphing to Solve a Quadratic Equation

Solve the quadratic equation by graphing:

x^2+6x=-5

Write the quadratic equation in standard form, where

a

represents the coefficient of

x^2

b

represents the coefficient of

x

, and

c

represents the last term.

ax^2+bx+c=0

Add 5 to both sides of the equation:

\begin{aligned}x^2+6x&={-5}\\{x^2}+6x+5&=0\end{aligned}

Write the related function of the quadratic equation:

\begin{aligned}f(x)&={ax^2+bx+c}\\{y}&={x^2}+6x+5\end{aligned}

Identify the values of

a

b

, and

c

\begin{aligned}a&={1}\\{b}&={6}\\{c}&=5\end{aligned}

Use the values of $a$ , $b$ , and $c$ to determine the vertex and axis of symmetry.

To determine the vertex, use the formulas for the

x

- and

y

-coordinates of the graph of a quadratic equation.

\begin{aligned}{\left(-\frac{b}{2a},c-\frac{b^2}{4a}\right)}&={\left(-\frac{6}{{2}\cdot{1}},5-\frac{6^2}{{4}\cdot{1}}\right)}\\&={\left(-\frac{6}{2},5-\frac{36}{4}\right)}\\&={(-3,5-9)}\\&=(-3,-4)\end{aligned}

To determine the location of the axis of symmetry, identify the

x

-coordinate of the vertex. The graph of the given equation has an axis of symmetry at

x=-3

Identify at least one other point on the graph by determining the

y

-intercept. Substitute zero for

x

\begin{aligned}y&=x^2+{6x+5}\\&=0^2+6(0)+{5}\\&=5\end{aligned}

The graph passes through the point

(0,5)

Another point that is easy to locate is 1 unit right of the vertex. From the vertex, move 1 unit right and $a$ units up (if $a$ is positive) or down (if $a$ is negative). Since $a=1$ , move 1 unit right and 1 unit up from the vertex. The graph passes through $(-2,-3)$ .

Use symmetry to identify points reflected across the axis of symmetry. The axis of symmetry is $x=-3$ , so the mirror images of the points $(0,5)$ and $(-2,-3)$ with respect to axis of symmetry are also points on the graph. Therefore, the graph passes through $(-6,5)$ and $(-4,-3)$ .

Plot the vertex, and draw the axis of symmetry. Then plot the additional points and sketch the graph, which is a parabola.

The $x$ -intercepts represent the solutions of the equation. Use the graph to identify the $x$ -intercepts.

The graph appears to pass through the points

(-5,0)

and

(-1,0)

, so the

x

-intercepts appear to be –5 and –1.

Verify the solutions by substituting the values of the

x

-intercepts in the original quadratic equation. Using the point

(-5, 0)

\begin{aligned}x^2+6x&={-5}\\(-5)^2+6(-5)&=-5\;\;\;\;\; \rm{?}\\25-30&=-5\;\;\;\;\;\rm{?}\\-5&=-5\;\;\;\;\;\checkmark\end{aligned}

Using the point

(-1, 0)

\begin{aligned}x^2+6x&={-5}\\(-1)^2+6(-1)&=-5 \;\;\;\;\;\rm{?}\\1-6&=-5\;\;\;\;\;\rm{?}\\-5&=-5\;\;\;\;\;\checkmark\end{aligned}

The solutions are

x=-5

and

x=-1

<Vocabulary >Factoring Quadratic Equations

Solving Quadratic Equations and Inequalities

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Graphing Quadratic Equations