Question

Graph the following parabolas by using intercepts, the vertex, and the axis of symmetry .

$y=16x^2+24x+9$

Short answer

The required graph is shown below:

Step-by-step solution

Given information

The given equation is$y=16x^2+24x+9$

Determine the axis of symmetry.

By comparing the given equation with the standard form $y=a{x}^2+bx+c$we get $a=16,b=24andc=9$.

Since the value of a is positive so parabola opens up.

The axis of symmetry is the vertical line:

$$\begin{gathered} x=-\frac{b}{2a} \\ =-\frac{24}{2\left(16\right)} \\ =-\frac{3}{4} \end{gathered}$$

The axis of symmetry is$x=-\frac{3}{4}$.

Determine the vertex.

Put $x=-\frac{3}{4}$into the given equation we get.

$$\begin{gathered} y=16{\left(-\frac{3}{4}\right)}^2+24\left(-\frac{3}{4}\right)+9 \\ =9-18+9 \\ =0 \end{gathered}$$

The vertex is$\left(-\frac{3}{4},0\right)$.

Determine the intercepts.

For y-intercept put $x=0$into the given equation.

$$\begin{gathered} y=16{\left(0\right)}^2+24\left(0\right)+9 \\ y=9 \end{gathered}$$

The y-intercept is $(0,9)$.

For x-intercept put $y=0$into the given equation.

role="math" localid="1653823101725" $$\begin{gathered} 0=16x^2+24x+9 \\ {(4x+3)}^2=0 \\ x=-\frac{3}{4} \end{gathered}$$

The x-intercept is $\left(-\frac{3}{4},0\right)$.

Draw the graph.

By using the above information the required graph is shown below: