answer the following for the given quadratic function.
f(x)=x^2+x^2-8
a. what is the vertex (h,k) of f?
(h,k)=? simplify your answer and type the answer in an ordered pair
b. what is the axis of symmetry?
a. (0,-8) is the vertex
b. axis of symmetry is zero.
I have a feeling there is a typo.
Why would anybody have two like terms in the equation ?
should it be
f(x) = x^2 + 2x - 8 ????
Yes, Reiny you caught the typo
It should read f(x)=x^2+2x-8, thank you for catching that. So, now my answers will change that Riana gave. Right?
yes,
complete the square ....
f(x) = x^2 + 2x +1 - 1 -8
= (x+1)^2 - 9
vertex is (-1,-9)
axis of symmetry: x = -1 or x+1 = 0
then your vertex is (-2,8)
and axis will be -1
the vertex can't be (-1,-9) its incorrect.
NO, check my reply
Yes it is !
http://www.wolframalpha.com/input/?i=plot+x%5E2+%2B+2x+-+8
To find the vertex (h, k) of a quadratic function, you can use the formula h = -b/2a and substitute it into the equation f(x) to find the corresponding value of k. Let's use this approach to answer part (a) of your question.
Given the quadratic function f(x) = x^2 + x^2 - 8, we can simplify it to f(x) = 2x^2 - 8.
Comparing it with the standard form of a quadratic function (f(x) = ax^2 + bx + c), we identify a = 2, b = 0, and c = -8.
Now, we can use the formula h = -b/2a to find the x-coordinate of the vertex. In our case, h = -0/2(2) = 0.
To find the y-coordinate (k) of the vertex, substitute the value of h back into the equation f(x): f(0) = 2(0)^2 - 8 = -8.
Therefore, the vertex of f(x) is (h, k) = (0, -8).
Moving on to part (b) of your question, the axis of symmetry is a vertical line that passes through the vertex, which has the equation x = h. In this case, x = 0.
So, the axis of symmetry for the quadratic function f(x) is x = 0.