Question

for problems 8 - 12: fill in the missing boxes using the provided information. equation $y=\frac{1}{4p}(x - h)^{2}+k$, $x=\frac{1}{4p}(y - k)^{2}+h$ vertex $(h,k)$ focus directrix axis of symmetry graph sketch 8 (3,7) (3,4) 9 (-2,5) $y = - 3$ 10 (4,10) $x = 9$ 11 (-8,-2) (-5,-2) 12 (-2,0) $y = 4$

Explanation:

Step1: Recall parabola properties

For a parabola of the form $y = \frac{1}{4p}(x - h)^2+k$ (vertical - axis) or $x=\frac{1}{4p}(y - k)^2+h$ (horizontal - axis), the vertex is $(h,k)$, the focus is $(h,k + p)$ for vertical - axis and $(h + p,k)$ for horizontal - axis, and the directrix is $y=k - p$ for vertical - axis and $x=h - p$ for horizontal - axis, and the axis of symmetry is $x = h$ for vertical - axis and $y = k$ for horizontal - axis.

Problem 8:

The vertex is $(h,k)=(3,7)$ and the focus is $(3,4)$. Since the $x$ - coordinates of the vertex and focus are the same, it is a vertical - axis parabola.

Step2: Calculate $p$

We know that for a vertical - axis parabola, if the vertex is $(h,k)$ and the focus is $(h,k + p)$, then $k + p=4$ and $k = 7$. So $7 + p=4$, which gives $p=- 3$.

Step3: Find the directrix

The directrix of a vertical - axis parabola is $y=k - p$. Substituting $k = 7$ and $p=-3$, we get $y=7-(-3)=10$. The axis of symmetry is $x = 3$.
The equation of the parabola is $y=\frac{1}{-12}(x - 3)^2+7$.

Problem 9:

The focus is $(-2,5)$ and the directrix is $y=-3$. Since the directrix is a horizontal line, it is a vertical - axis parabola.

Step1: Calculate the vertex

The vertex $(h,k)$ of a vertical - axis parabola is the mid - point between the focus and the point on the directrix directly below (or above) the focus. The $x$ - coordinate of the vertex is $h=-2$. The $y$ - coordinate of the vertex $k=\frac{5+( - 3)}{2}=1$.

Step2: Calculate $p$

The distance from the vertex $(h,k)=(-2,1)$ to the focus $(-2,5)$ is $p = 4$. The axis of symmetry is $x=-2$. The equation of the parabola is $y=\frac{1}{16}(x + 2)^2+1$.

Problem 10:

The vertex is $(4,10)$ and the directrix is $x = 9$. Since the directrix is a vertical line, it is a horizontal - axis parabola.

Step1: Calculate $p$

For a horizontal - axis parabola with vertex $(h,k)=(4,10)$ and directrix $x=h - p$, we have $4 - p=9$, so $p=-5$.

Step2: Find the focus

The focus of a horizontal - axis parabola is $(h + p,k)$. Substituting $h = 4$, $p=-5$ and $k = 10$, we get the focus $(-1,10)$. The axis of symmetry is $y = 10$. The equation of the parabola is $x=\frac{1}{-20}(y - 10)^2+4$.

Problem 11:

The vertex is $(-8,-2)$ and the focus is $(-5,-2)$. Since the $y$ - coordinates of the vertex and focus are the same, it is a horizontal - axis parabola.

Step1: Calculate $p$

For a horizontal - axis parabola, if the vertex is $(h,k)=(-8,-2)$ and the focus is $(h + p,k)$, then $h + p=-5$ and $h=-8$. So $-8 + p=-5$, which gives $p = 3$.

Step2: Find the directrix

The directrix of a horizontal - axis parabola is $x=h - p$. Substituting $h=-8$ and $p = 3$, we get $x=-11$. The axis of symmetry is $y=-2$. The equation of the parabola is $x=\frac{1}{12}(y + 2)^2-8$.

Problem 12:

The focus is $(-2,0)$ and the directrix is $y = 4$. Since the directrix is a horizontal line, it is a vertical - axis parabola.

Step1: Calculate the vertex

The vertex $(h,k)$ is the mid - point between the focus and the point on the directrix directly above the focus. The $x$ - coordinate of the vertex is $h=-2$. The $y$ - coordinate of the vertex $k=\frac{0 + 4}{2}=2$.

Step2: Calculate $p$

The distance from the vertex $(-2,2)$ to the focus $(-2,0)$ is $p=-2$. The axis of symmetry is $x=-2$. The equation of the parabola is $y=\frac{1}{-8}(x + 2)^2+2$.

Equation	Vertex $(h,k)$	Focus	Directrix	Axis of Symmetry	Graph Sketch (description)

Answer:

Equation	Vertex $(h,k)$	Focus	Directrix	Axis of Symmetry	Graph Sketch (description)
$y=\frac{1}{16}(x + 2)^2+1$	$(-2,1)$	$(-2,5)$	$y=-3$	$x=-2$	Up - opening parabola at $(-2,1)$
$x=\frac{1}{-20}(y - 10)^2+4$	$(4,10)$	$(-1,10)$	$x = 9$	$y = 10$	Left - opening parabola at $(4,10)$
$x=\frac{1}{12}(y + 2)^2-8$	$(-8,-2)$	$(-5,-2)$	$x=-11$	$y=-2$	Right - opening parabola at $(-8,-2)$
$y=\frac{1}{-8}(x + 2)^2+2$	$(-2,2)$	$(-2,0)$	$y = 4$	$x=-2$	Down - opening parabola at $(-2,2)$