What is the maximum value of the function ƒ(x) = –x2 + 6x + 10?

The simplest way to identify the vertex of a quadratic equation is to first use the formula -b / 2a

Quadratic equation equations are in the form of
ax^2 + bx + c
Your function is
-1x^2 + 6x + 10

What you want to do is identify the a, b and c values, and then substitute the a and b values in the formula.

As you can see, the a value is -1, and the b value is 6.

We substitute these numbers into our formula, and then

-b / 2a becomes

-(6) / 2 (-1)
-6 / -2 = 3

Therefore, our x coordinate of the vertex is 3.

Now, what you want to do is substitute 3 into the original function in order to get the y coordinate of the vertex.

ƒ(x) = –x2 + 6x + 10?
f (3) = -(3)^2 + 6(3) + 10
f (3) = -9 + 18 + 10
f (3) = 19

The y coordinate of the vertex is 19.

The maximum value, or vertex of the function is (3, 19)

Hope this helps!!
Let me know if there's something you don't understand and I'll try to explain as best I can.

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