Chapter 7.5, Problem 14OE

a)

To determine

To determine whether the parabola opens upward or downward.

Answer to Problem 14OE

Parabola opens upward

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Explanation of Solution

Given information :

Given equation $y+2=5{\left(x+1\right)}^2$

We know that the graph of the equation $y-k=a{\left(x-h\right)}^2$ opens upwards when $a>0$ and downwards when $a<0$ .

The coordinates of the vertex is given by $\left(h,k\right)$

Axis of symmetry is given by the value of $h$

Now we compare the given equation with the standard equation, we get,

Here the value of $a=5,\left(a>0\right)$ hence the graph of parabola opens upwards.

Axis of symmetry is $x=-1$

b)

To determine

The coordinates of the vertex.

Answer to Problem 14OE

Coordinates of the vertex is (-1, -2)

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Explanation of Solution

Given information :

Given equation $y+2=5{\left(x+1\right)}^2$

We know that the graph of the equation $y-k=a{\left(x-h\right)}^2$ opens upwards when $a>0$ and downwards when $a<0$ .

The coordinates of the vertex is given by $\left(h,k\right)$

Axis of symmetry is given by the value of $h$

Now we compare the given equation with the standard equation, we get,

. $k=-2$

And $h=-1$

Therefore the coordinates of the vertex is (-1, -2)

c)

To determine

The equation of the axis of symmetry.

Answer to Problem 14OE

Axis of symmetry is $x=-1$

.

Explanation of Solution

Given information :

Given equation $y+2=5{\left(x+1\right)}^2$

We know that the graph of the equation $y-k=a{\left(x-h\right)}^2$ opens upwards when $a>0$ and downwards when $a<0$ .

The coordinates of the vertex is given by $\left(h,k\right)$

Axis of symmetry is given by the value of $h$

Now we compare the given equation with the standard equation, we get,

$k=-2$

And $h=-1$

Therefore the coordinates of the vertex is (-1, -2)

Axis of symmetry is $x=-1$