Chapter 10.2, Problem 48AYU

In Problems 39-56, find the vertex, focus, and directrix of each parabola. Graph the equation by hand. Verify your graph using a graphing utility.

$x^2+\text{ 6}x-\text{ 4}y\text{ + 1 = 0}$

To determine

(a & b)

To find: Vertex, focus and directrix of the parabola given by, $x²+6x-4y+1=0$.

Answer to Problem 48AYU

Vertex: $(-3,-2)$.

Focus: $\left(-3, -1\right)$.

Directrix D: $y=-3$.

Explanation of Solution

Given:

$x²+6x-4y+1=0$

Formula used:

Equation	Vertex	Focus	Directrix	Description
${\left(x-h\right)}^2=4a\left(y-k\right)$	$(h,k)$	$\left(h,k+a\right)$	$y=k-a$	Parabola, axis of symmetry is parallel to $y\text{-axis}$, opens up.

Calculation:

The equation $x²+6x-4y+1=0$.

${\left(x+3\right)}^2=4\left(y+2\right)$, is of the form ${\left(x-h\right)}^2=4a\left(y-k\right)$.

$h=-3;k=-2$

Hence, $4a=4;a=1$.

Therefore, vertex: $\left(h,k\right) =\left(-3,-2\right)$.

Focus: $\left(h, k+a\right)=\left(-3, -1\right)$.

Directrix D: $y=k-a=-2-1=-3;y=-3$.

Parabola, axis of symmetry is parallel to $y\text{-axis}$, and it opens up.

To determine

c.

To graph: The parabola for the equation $x²+6x-4y+1=0$ by hand verify using a graphing tool.

Answer to Problem 48AYU

Explanation of Solution

Given:

$x²+6x-4y+1=0$

Calculation:

The graph of the parabola $x²+6x-4y+1=0$.

$x²+6x-4y+1=0$

By using the table above, we obtain few points $\left(0,.25\right),\left(-1,-1\right),\left(-5,-1\right)$ and $\left(-6,0.25\right)$. We plot the graph of $x²+6x-4y+1=0$ using these points and the graph opens up. The points are shown above in the graph.

Using graphing utility we verify the graph.

The Vertex $\text{V}\left(-3, -2\right)$, focus $\text{F}\left(-3, -1\right)$ is shown in the graph above. The directrix D: $y=-3$ is shown in the dotted line.