# Which of the following equations is of a parabola with a vertex at (0, -5)? y = (x - 5)2 y = (x + 5)2 y = x2 - 5 y = x2

###### Question:

## Answers

Put 0 for x in each possible choice and see which one gives you y = -5.

The appropriate choice is

y = x² - 5

Of course, you know the vertex form is

y = a(x -h)² + k

for vertex (h, k) and scale factor "a".

Then for (h, k) = (0, -5), this is

y = ax² -5

Only one choice matches that: y = x² -5. (for a=1)

The equation

[tex]y = {x}^{2} - 5[/tex]

will have a parabola with a vertex at (0,-5) as the y intercept is - 5. Remember, quadratic equations can often be written in the form

[tex]y = a {x}^{2} + bx + c[/tex]

Where a is the coefficient of the quadratic, b is the coefficient of the multiple and C is the vertex/y-intercept.

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Step-by-step explanation:

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[tex]\bf ~~~~~~\textit{parabola vertex form} \\\\ y=a(x- h)^2+ k \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k}) \\\\[-0.35em] ~\dotfill\\\\ y=(x-5)^2\implies y=(x-\stackrel{h}{5})^2+\stackrel{k}{0}~~\checkmark \\\\\\ y=(x+5)^2\implies y=[x-(-5)]^2\implies y=[x-(-\stackrel{h}{5})]^2+\stackrel{k}{0}~~\checkmark[/tex]

Equation of parabola is y = x² - 5 .

Step-by-step explanation:

Given : parabola with a vertex at (0, -5).

To find : which of the following equations is of a parabola.

Solution : We have given vertex at (0, -5).

Vertex form of parabola y = a(x - h)² + k .

Where (h ,k ) is vertex

We have h = 0 , k = -5 .

Equation : y = 1(x - 0)² - 5

y = x² - 5 .

Therefore, Equation of parabola is y = x² - 5 .

The equation that has a vertex in (0, -5) is the first one.

Step-by-step explanation:

A standard form second degree equation is given by the following expression:

[tex]y(x) = a*x^2 + b*x + c[/tex]

For which the vertex coordinates can be calculated by:

[tex]x_{vertex} = -\frac{b}{2*a}[/tex]

While "y" for the vertex can be found by applying this coordinate on the expression. Using this knowledge in each equation gives us:

1. [tex]y(x) = x^2 - 5[/tex]

[tex]a = 1\\b = 0\\c = -5[/tex]

Therefore the vertex coordinate is:

[tex]x_{vertex} = -\frac{b}{2*a} = -\frac{0}{2*1} = 0[/tex]

[tex]y_{vertex} = 0^2 - 5 = -5[/tex]

This parabola has a vertex in (0,-5).

2. [tex]y(x) = x^2 + 5[/tex]

[tex]a = 1\\b = 0\\c = 5[/tex]

Therefore the vertex coordinate is:

[tex]x_{vertex} = -\frac{b}{2*a} = -\frac{0}{2*1} = 0[/tex]

[tex]y_{vertex} = 0^2 + 5 = 5[/tex]

This parabola has a vertex in (0,5).

3. [tex]y(x) =(x+ 5)^2 = x^2 + 10*x + 25[/tex]

[tex]a = 1\\b = 10\\c = 25[/tex]

Therefore the vertex coordinate is:

[tex]x_{vertex} = -\frac{b}{2*a} = -\frac{10}{2*1} = -5[/tex]

[tex]y_{vertex} = (-5)^2 + 10*(-5) + 25 = 0[/tex]

This parabola has a vertex in (-5,0).

4. [tex]y(x) =(x- 5)^2 = x^2 - 10*x + 25[/tex]

[tex]a = 1\\b =- 10\\c = 25[/tex]

Therefore the vertex coordinate is:

[tex]x_{vertex} = -\frac{b}{2*a} = -\frac{-10}{2*1} = 5[/tex]

[tex]y_{vertex} = (5)^2 - 10*(5) + 25 = 0[/tex]

This parabola has a vertex in (5,0).

The x, which is zero in this case will always be the number that goes on the inside of the equation, so far you have y=(x+0)^2 or y=x^2.

The y, which is negative will go on the end of the equation. So you get y=x^2-5

The equation is y=x^2-5 or C if it is multiple choice

The equation of a parabola with Vertex V=(h,k) is

y=a(x-h)^2+k

If the Vertex is V=(0,-5)=(h,k)→h=0, k=-5

Replacing h=0 and k=-5 in the equation above:

y=a(x-0)^2+(-5)

y=ax^2-5

Third option: y=x^2-5