Q1. - Algebra 2

Question

Identify the vertex, focus, axis of symmetry and directrix of the graph of the parabola $y=4{\left(x-3\right)}^2-7$.

Step-by-Step Solution

Answer

The vertex is $\left(3,-7\right)$. Focus is $\left(3,-\frac{111}{16}\right)$. Axis of symmetry is $x=3$. The equation of directrix is $y=-\frac{113}{16}$.

Step 1. Write down the given information.

The given equation is $y=4{\left(x-3\right)}^2-7$.

Step 2. Concept used.

For two different forms of equations of parabola stated below, use the following key-concept to find vertex, axis of symmetry, focus, directrix, direction of opening of parabola and length of latus rectum.

Step 3. Comparing the given equation with standard form.

On comparing the equation $y=4{\left(x-3\right)}^2-7$ with standard form $y=a{\left(x-h\right)}^2+k$, we get $a=4,h=3\text{ and }k=-7$.

Step 4. Evaluating vertex, focus, equations of axis of symmetry and directrix of parabola.

The vertex and focus, the equations of the axis of symmetry and directrix of the parabola are evaluated as:

The vertex is $\left(3,-7\right)$.

Focus is evaluated as:

 $\begin{array}{c}\left(h,k+\frac{1}{4a}\right)=\left(3,-7+\frac{1}{4\left(4\right)}\right)\\ =\left(3,-\frac{111}{16}\right)\end{array}$

Axis of symmetry is $x=3$.

The equation for directrix is evaluated as:

 $\begin{array}{c}y=k-\frac{1}{4a}\\ y=-7-\frac{1}{4\left(4\right)}\\ y=-\frac{113}{16}\mathrm{....}\left(\text{Directrix}\right)\end{array}$

Step 5. Conclusion.

The vertex is $\left(3,-7\right)$. Focus is $\left(3,-\frac{111}{16}\right)$. Axis of symmetry is $x=3$. The equation of directrix is $y=-\frac{113}{16}$.