91Ó°ÊÓ

$$\begin{gathered} f\left(x\right)=a{x}^2+bx+c \\ f\left(x\right)=5x^2+10x+3 \end{gathered}$$

Since a is 5 , the parabola opens upward.

$$\begin{gathered} x=-\frac{b}{2a} \\ x=-\frac{10}{2\left(5\right)} \\ x=-1 \end{gathered}$$

The axis of symmetry is $x=-1$

Find $f(-1)$

$$\begin{gathered} f\left(x\right)=5x^2+10x+3 \\ f(-1)=5{(-1)}^2+10(-1)+3 \\ f(-1)=5-10+3 \\ f(-1)=-2 \end{gathered}$$

The vertex is$(-1,-2)$

The y-intercept occurs when x = 0 . Find f (0) .

$$\begin{gathered} f\left(x\right)=5x^2+10x+3 \\ f\left(0\right)=5{\left(0\right)}^2+10\left(0\right)+3 \\ f\left(0\right)=3 \end{gathered}$$

The y-intercept is (0,3)

The point $(0,3)$is one unit to the right of the line of symmetry.

The point one unit to the left of the line of symmetry is$(-2,3)$

Point symmetric to the y-intercept is$(-2,3)$

The x-intercept occurs when f (x) = 0

Find f (x) = 0 .

$$\begin{gathered} f\left(x\right)=5x^2+10x+3 \\ 0=5x^2+10x+3 \end{gathered}$$

Use the Quadratic Formula:$x=\frac{-bÂ\pm \sqrt{b^2-4ac}}{2a}$

Substitute in the values of a, b, and c. role="math" localid="1645204226342" $$\begin{gathered} x=\frac{-10Â\pm \sqrt{{10}^2-4×5×3}}{2\left(5\right)} \\ x=\frac{-10Â\pm \sqrt{100-60}}{10} \\ x=\frac{-10Â\pm \sqrt{40}}{10} \\ x=\frac{-10Â\pm 2\sqrt{10}}{10} \\ x=\frac{-5Â\pm \sqrt{10}}{5} \end{gathered}$$

Write as two equations:

$$\begin{gathered} x=\frac{-5+\sqrt{10}}{5};x=\frac{-5-\sqrt{10}}{5} \\ x≈-0.368;x≈-1.632 \end{gathered}$$

The approximate values of the

x-intercepts are$(-0.368,0)and(−1.632,0)$

Graph the parabola using the points found.