Question

Please help solve with walk through. Identify the vertex and axis of symmetry of the function y=4(x+3)^2+2.

Answer 1

the vertex is (-3,2)

Answer 2

the axis of symmetry is x = -3

Answer 3

I'd strongly recommend looking up or verifying the equation of a parabola with vertex at (h,k). It's y = a(x-h)^2 + k, where h is the x-coordinate of the vertex and k is the y-coordinate.

Compare the given equation, y=4(x+3)^2+2, to this general equation, y=a(x-h)^2+k.
Here you can identify a=4, h=-3, and k=+2.

Thus, the vertex is at (-3,2). Since your parabola is a vertical one, the axis of symmetry passes through the vertex and has the equation x = -3.

Hope this discussion makes it easier for you to do similar problems on your own.

Best of luck to you!

Answer 4

thanks man could you help me with one more

Answer 5

Whatcha got up your sleeve?

Answer 6

its y=-3x^2+18x-24 whats the vertex and AOS

Answer 7

Hi, Max,
This problem is somewhat similar to the preceding one, but is complicated by the fact that x shows up in TWO terms here, whereas it showed up only ONCE in the previous question.

We still want to end up with an equation for the parabola that looks like y = a(x-h)^2 + k.

To do this, we need to apply a technique called "completing the square". Have you heard of and/or used that before?

Answer 8

i have not

Answer 9

Without going into a lot of detail up front, I'll just walk you through the necessary steps of this "completing the square" procedure.

y=-3x^2+18x-24 can be re-written as y = -3(x^2 -6x ) - 24. Be certain that you understand and can verify this.

Answer 10

ok yeah ive done that i didnt know it was called that

Answer 11

wouldnt it be -3(x^2-6x+8)

Answer 12

Now we want to "complete the square" in (x^2 - 6x ).
Take half of the coefficient of x: in other words, take (1/2)(-6) = -3.
Square that result, then add and subtract the same number to x^2 - 6x.
What results is a quadratic which is a perfect square, less 9:

Answer 13

I appreciate your trying to solve this. While factorable, your (x^2-6x+8) isn't a perfect square.
Try this:
y = - 3(x^2-6x+9 -9) -24 (Can you verify this?)

Answer 14

Yeah i understand now

Answer 15

Rewrite that perfect square as the square of a binomial:
y= -3(x-3)^2 +27 -24,
or
y =-3(x-3)^2 + 3
y = a(x-h)^2 + k
Thus, h = 3 and k = 3, and so the vertex is at (h,k): (3,3).
What is the axis of symmetry?

Answer 16

aos=3?

Answer 17

Yes, but it's best to write that as an equation of a straight line: x = 3.

Answer 18

That concludes the solution.
Great working with you.
Best of luck to you.

Answer 19

ok thanks man you helped me alot I am studying for a midterm tomorrow.

Answer 20

Hope it goes extremely well for you. Try practicing these procedures if you have the time and energy left. Merry Christmas. Over and out.