**quadratic equation**is a polynomial equation, where $a$, $b$, and $c$ are real numbers and $a\neq0$, that can be written in the form:

**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:

**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

**is the value of $x$, where the graph touches or crosses the $x$-axis.**

*x*-interceptTo 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.

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. |

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

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 $(-4,-3)$.

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 $(-5, 0)$: