SOLUTION: what is the vertex of f(x)=x^2-8x-5 f(x)=-x^2+10x+6 f(x)=3x^2-12x+9 f(x)=-2x^2+2x+7 f(x)

Question 478190: what is the vertex of f(x)=x^2-8x-5
f(x)=-x^2+10x+6
f(x)=3x^2-12x+9
f(x)=-2x^2+2x+7
f(x)=8-x^2
I have bee trying to solve and find the vertexes for these five problems since last Sunday. I don't understand quadratic equations at all.
Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
The vertex of a quadratic function is the turning point of the function.
At this point, the slope of the function, dy/dx = 0.
The equation for the function in standard form is:
y = ax^2 + bx + c
To find the vertex, compute dy/dx and set equal to 0:
dy/dx = 0 = 2ax + b
Solving for x gives x = -b/2a
Now substitute this value back in the original equation to get the y-value of the vertex:
y = a(-b/2a)^2 + b(-b/2a) + c
This gives y = c - b^2/(4a)
We can use this information to find the vertices of the given functions.
e.g. f(x) = x^2-8x-5
The vertex is x = 8/2 = 4; y = -5 - (-8^2/4) -> y = -21
I will leave it to you to compute the rest!
Ans: (4,-21)
We can graph the function to check our answer

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