Question

PLEASE HELP WILL GIVE BRAINLY!!!!!!!!!What is the vertex of the graph of the function f(x) = x2 + 8x − 2 ?

Answer 1

Answer: D.  (-4, -18)

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Explanation:

A graph is a nice addition (and it will likely help us see the vertex directly), but it isn't necessary because we can use the equation given to us. Though I recommend using a graphing calculator to confirm the answer.

The original function is the same as y = 1x^2+8x + (-2). We see that it is in the form y = ax^2+bx+c where

a = 1

b = 8

c = -2

Use the values of 'a' and b to get the value of h, which is the x coordinate of the vertex.

h = -b/(2a)

h = -8/(2*1)

h = -4

The x coordinate of the vertex is x = -4. Plug this into the original equation to get

f(x) = x^2+8x-2

f(-4) = (-4)^2 + 8(-4) - 2

f(-4) = -18

Plugging x = -4 into f(x) leads to y = -18. The point (-4, -18) is on the parabola. Furthermore, this is the vertex (h,k)

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Alternatively, you can complete the square as shown below

y = x^2 + 8x - 2

y = (x^2 + 8x) - 2

y = (x^2 + 8x + 0) - 2

y = (x^2 + 8x + 16 - 16) - 2

y = (x^2 + 8x + 16) - 16 - 2

y = (x+4)^2 - 18

y = 1(x+4)^2 - 18

y = 1(x-(-4))^2 - 18

The last equation is in the form y = a(x-h)^2 + k with (h,k) = (-4,-18) being the vertex. The 16 is the result of taking half of 8 and squaring that result. We have 16-16 = 0 to make sure that we don't change the equation and keep things balanced. This is the same as adding 16 to both sides. All of this is done so we can end up with the (x+4)^2 perfect square portion.You can expand out (x+4)^2 - 18 and you should get x^2+8x-2 again.

Answer 2

D is the answer

Because you need to rewrite in vertex form and use this form to find the vertex (h,k)

Which should get you to

(-4,-18)