A parabola whose equation is y = (x - 5)^2 - 2 has a vertex at _____.

2 answers:

7 0

(5, -2) because y=(x-5)^2-2 is really y=x^2-10x+23 after you foil then take the derivative which is y'=2x-10 solve for x and you get 5 then put 5 in the original equation and you get -2 so the answer is (5, -2)<span />

omeli [17]3 years ago

6 0

Answer:

Option 3 - (5,-2)

Step-by-step explanation:

Given : A parabola whose equation is $y=(x-5)^2-2$

To find : The vertex of the parabola

Solution :

The general vertex form of parabola is

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

Where, (h,k) is the vertex of the function.

The given function is already in vertex form so we compare it with general form

$y=(x-5)^2-2$

a =1, h=5 and k=-2

So, The vertex of the given parabola is (h,k)=(5,-2)

Therefore, Option 3 is correct.

You might be interested in

The coordinates of the vertices of quadrilateral ABCD are A(−4, −1) , B(−1, 2) , C(5, 1) , and D(1, −3) .

finlep [7]

Hi there, KawallPotato! :)

Finding the slope of a line is finding the rise over run, or the change in y over the change in x. Let me show you what I mean.

Slope = change in y / change in x

Or in other words, $\frac{y2 - y1}{x2 - x1}$

Let's try it with AB. The first thing to consider is the placement of the letters. The blank space is asking for the slope of point A to point B, which means that the coordinates of A are your y₁ and x₁ values and the coordinates of B are your y₂ and x₂ values.

Put them in the equation like this:

$\frac{2 - (-1)}{-1 - (-4)}$