In the following exercises, graph by using intercepts, the vertex, and the axis of symmetry.

$y=x^2+8x+15$

$$\begin{gathered} y=a{x}^2+bx+c \\ y=x^2+8x+15 \end{gathered}$$

The value of coefficient ais positive therefore we can say that the parabola opens up.

The axis of symmetry is the line $x=-\frac{b}{2a}$.

Substitute the values of a, and binto the equation.

$$\begin{gathered} x=-\frac{8}{2\left(1\right)} \\ =-4 \end{gathered}$$

Therefore, the axis of symmetry is$x=-4$.

The vertex is on the line of symmetry, so its x-coordinate will be$x=-4$.

Now substitute the value of xinto the equation,

$$\begin{gathered} y=x^2+8x+15 \\ ={\left(-4\right)}^2+8\left(-4\right)+15 \\ =16-32+15 \\ =-1 \end{gathered}$$

Therefore, the vertex is$\left(-4,-1\right)$.

Now substitute xequal to 0 in the equation to find the intercept y,

$$\begin{gathered} y=0^2+8\left(0\right)+15 \\ =15 \end{gathered}$$

Therefore, the point $\left(0,15\right)$ is the y-intercept.

We can see that the resultant point is right of the line of symmetry by 4 units.

Therefore the point left to the line of symmetry by 4 units is role="math" localid="1653914782016" $\left(-8,15\right)$.

Now substitute yequal to 0 in the equation to find the intercept x,

role="math" localid="1653915035105" $$\begin{gathered} x^2+8x+15=0 \\ {x}^2+5x+3x+15=0 \\ x\left(x+5\right)+3\left(x+5\right)=0 \\ \left(x+3\right)\left(x+5\right)=0 \\ x=-3,-5 \end{gathered}$$

Therefore, the points of the x-intercept are$\left(-3,0\right)\mathrm{and}\left(-5,0\right)$.