Q70. - Algebra 1

Question

CHALLENGE Using the axis of symmetry and one x-intercept, write an equation for the graph shown.

Step-by-Step Solution

Answer

The equation for given graph is $y=-x^2+\mathbf{6}x+\mathbf{16}$.

Step 1. Axis of symmetry.

From the graph, it can be observed that the vertex of a parabola is $\left(3,25\right)$. Therefore, the axis of symmetry is $x=3$. 

Step 2. Substitution.

Substitute the value of x in the equation, $x=-\frac{b}{2a}$.

$$\begin{gathered} 3=-\frac{b}{2a} \\ b=-6a \end{gathered}$$

Step 3. Y-intercept and x-intercept.

From the graph, it can be observed that the y-intercept is $c=16$. The parabola has x-intercept at $\left(-2,0\right)$ and $\left(8,0\right)$.

Step 4. Substitution.

Substitute $\left(8,0\right)$ for $\left(x,y\right)$, 16 for c and $-6$ for b into $y=a{x}^2+bx+c$.

$$\begin{gathered} a{\left(8\right)}^2+\left(-6a\right)\left(8\right)+16=0 \\ 64a-48a+16=0 \\ 16a=-16 \\ a=-\frac{16}{16} \\ =-1 \end{gathered}$$

 

Therefore, the value of b is

$$\begin{gathered} b=-6a \\ =-6\left(-1\right) \\ =6 \end{gathered}$$

Step 5. Write equation for given graph.

Substitute $-1$ for a, 6 for b and 16 for c in the standard form of quadratic equation.

$$\begin{gathered} y=a{x}^2+bx+c \\ =-x^2+6x+16 \end{gathered}$$

 

Therefore, the equation for given graph is $y=-x^2+6x+16$.