Question

In the following exercises, graph each equation by using properties.

$y=8x^2-10x+3$

Short answer

On graphing the equation we get,

Step-by-step solution

Step 1. Given information.

Consider the given function,

$y=8x^2-10x+3$

Compare $y=a{x}^2+bx+c$ with the given function,

$$\begin{gathered} a=8, \\ b=-10, \\ c=3 \end{gathered}$$

Since $a=8$, the parabola opens to the right.

The axis of symmetry, $x=-\frac{b}{2a}$. Then,

$$\begin{gathered} x=-\frac{\left(-10\right)}{2\left(8\right)} \\ =\frac{5}{8} \end{gathered}$$

Step 2. Find the vertex.

The vertex is on the line $x=\frac{5}{8}$.

Substitute $x=\frac{5}{8}$in the original equation,

$$\begin{gathered} y=8{\left(\frac{5}{8}\right)}^2-10\left(\frac{5}{8}\right)+3 \\ =\frac{25}{8}-\frac{25}{4}+3 \\ =-\frac{1}{8} \end{gathered}$$

Therefore, the vertex is $\left(\frac{5}{8},-\frac{1}{8}\right)$.

Substitute $y=0$in the original equation,

$$\begin{gathered} 0=8x^2-10x+3 \\ x=\frac{3}{4},\frac{1}{2} \end{gathered}$$

Therefore, the x-intercepts are$\left(\frac{3}{4},0\right),\left(\frac{1}{2},0\right)$.

Step 3. Find the y-intercept.

Substitute $x=0$in the original equation,

$$\begin{gathered} y=8{\left(0\right)}^2-10\left(0\right)+3 \\ y=3 \end{gathered}$$

Therefore, the y-intercept is $\left(0,3\right)$.

The point symmetric to localid="1645199207623" $\left(0,3\right)$across the axis of symmetry is $\left(\frac{5}{4},3\right)$.

On graphing the parabola we get,