Chapter 10.2, Problem 15AYU

In Problems 13-20, the graph of a parabola is given. Match each graph to its equation.

(A) $y^2\text{ = 4}x$     (C) $y^2\text{ = }-\text{4}x$     (E) ${\left(y-\text{ 1}\right)}^2\text{ = 4}\left(x-\text{ 1}\right)$     (G) ${\left(y-\text{ 1}\right)}^2\text{ = }-\text{4}\left(x-\text{ 1}\right)$

(B) $x^2\text{ = 4}y$     (D) $x^2\text{ = }-\text{4}y$     (F) ${\left(x+\text{ 1}\right)}^2\text{ = 4}\left(y+\text{ 1}\right)$     (H) ${\left(x+\text{ 1}\right)}^2\text{ = }-\text{4}\left(y+\text{ 1}\right)$

To determine

The equation from the following that matches with the graph.

A. $y^2=4x$   B. $x^2=4y$   C. $y^2=-4x$   D. $x^2=-4y$

E. ${\left(y-1\right)}^2=4\left(x-1\right)$   F. ${\left(x+1\right)}^2=4\left(y+1\right)$

G. ${\left(y-1\right)}^2=-4\left(x-1\right)$   H. ${\left(x+1\right)}^2=-4\left(y+1\right)$

Answer to Problem 15AYU

H. ${\left(x+1\right)}^2=-4\left(y+1\right)$

Explanation of Solution

Given:

Formula used:

Vertex	Focus	Directrix	Equation	Description
$\left(h,k\right)$	$\left(0,k-a\right)$	$y=k+a$	${\left(x-h\right)}^2=-4a\left(y-k\right)$	Parabola, axis of symmetry is parallel to $y\text{-axis}$, opens down

Calculation:

The graph opens down, the vertex is at $V\left(-1,-1\right)$, the axis of symmetry is parallel to $y\text{-axis}$. The equation for the parabola which opens down is given by,

${\left(x-h\right)}^2=-4a\left(y-k\right)$

${\left(x+1\right)}^2=-4a\left(y+1\right)$

The given graph has equation of the form ${\left(x+1\right)}^2=-4a\left(y+1\right)$ and it matches with (H).