What is the equation of the parabola that has its vertex at [tex]$(8,-1)$[/tex] and a [tex]$y$[/tex]-intercept of [tex]$(0,-17)$[/tex]?

A) [tex]y = -\frac{1}{4}(x-8)^2 - 1[/tex]

B) [tex]y = \frac{1}{4}(x-8)^2 - 1[/tex]

C) [tex]y = -\frac{1}{4}(x-8)^2 + 1[/tex]

D) [tex]y = -\frac{1}{4}(x+8)^2 - 1[/tex]



Answer :

To find the equation of the parabola with a given vertex at [tex]\((8, -1)\)[/tex] and a [tex]\(y\)[/tex]-intercept at [tex]\((0, -17)\)[/tex], we should use the vertex form of a parabola equation, which is given by:

[tex]\[ y = a(x - h)^2 + k \][/tex]

Here, [tex]\((h, k)\)[/tex] is the vertex of the parabola. Substituting the vertex [tex]\((8, -1)\)[/tex] into the equation, we get:

[tex]\[ y = a(x - 8)^2 - 1 \][/tex]

Next, we need to use the [tex]\(y\)[/tex]-intercept to find the value of [tex]\(a\)[/tex]. The [tex]\(y\)[/tex]-intercept is the point where [tex]\(x = 0\)[/tex] and [tex]\(y = -17\)[/tex]. Substituting [tex]\(x = 0\)[/tex] and [tex]\(y = -17\)[/tex] into the equation, we obtain:

[tex]\[ -17 = a(0 - 8)^2 - 1 \][/tex]

Simplifying the equation:

[tex]\[ -17 = a(64) - 1 \][/tex]

Add 1 to both sides:

[tex]\[ -16 = 64a \][/tex]

Solving for [tex]\(a\)[/tex]:

[tex]\[ a = \frac{-16}{64} = -\frac{1}{4} \][/tex]

Now that we have the value of [tex]\(a\)[/tex], we can write the final equation of the parabola:

[tex]\[ y = -\frac{1}{4}(x - 8)^2 - 1 \][/tex]

Thus, the equation of the parabola is:

[tex]\[ y = -\frac{1}{4}(x - 8)^2 - 1 \][/tex]

This corresponds to option:
A) [tex]\( y = -\frac{1}{4}(x - 8)^2 - 1 \)[/tex]