Question

Write the function f(x)= 2x^2+8x+7 in the form of f(x)= a(x-h)^2 + k and find the vertex?

Answer 1

Start by expanding your target form:

a(x-h)^2 + k
= a(x^2 - 2xh + h^2) + k, form of the target

What you have:

2x^2+8x+7

How can you make that look like the expanded form of the target?

Well, the target has the coefficient a on the whole expanded square, but has no coefficient on x^2 (or really a silent coeffcient of 1)

So you can say let a=2 and factor 2 out of your expression

2(x^2+4x+7/2)

now your term in x is +4x, but the x term in the target is -2hx

So what for what value of h is +4x = -2hx ?

This is easy to solve for h: +4x/(-2x) = h; -2 = h

So now you can rewrite your expression as

2(x^2-2(-2)x+7/2)

and so far we know a=2 and h=-2

Comparing with the expanded form of the target:

a(x^2 - 2xh + h^2) + k,

we see we need the the third term of the expression to be h^2, or since we have h=-2, the third term needs to be 4. but what we have is 7/2. We need to add 1/2 to it to make 4, but we have to take 1/2 away somehow to balance that:

2(x^2-2(-2)x+7/2 + 1/2 -1/2)

2(x^2-2(-2)x+8/2 -1/2)

2(x^2-2(-2)x+4 -1/2)

Now let's kick that nasty -1/2 out of the parenthesis, but to do so, we have to double it, since everything in the parenthesis is multiplied by 2 (that a we were looking for)

2(x^2-2(-2)x+4) -1

Now the stuff in the parenthesis is a perfect square, so let's express it as such.

2(x-(-2))^2 -1

Now let's compare with our original target:

a(x-h)^2 + k

whoa! if a=2, h=-2, and k=1, they are the same!

f(x)= 2x^2+8x+7=2(x-(-2))^2 -1

and the vertex when a parabola is expressed this way is (h,k), in this case (-2,1)