Question

Find the maximum y-value on the graph of y=f(x)
f(x)= -x^2+4x+9

Answer 1

okay finally some calc1

Answer 2

i've got a couple of questions too!

Answer 3

y = -x^2 + 4x + 9
to maximize that take the derivative and set that equal to 0

Answer 4

y' = -2x + 4 = 0
-2x + 4 = 0
-2x = -4
x = 2

Answer 5

Soo when x = 2, y = -2^2 + 4*2 + 9 = -4 + 8 + 9 = 4+9 = 13

Answer 6

Would or could you help with more of my questions?

Answer 7

@bahrom7893

Answer 8

sure go ahead, ask away bud

Answer 9

make me feel like i actually know something lol

Answer 10

for y=x^2-1 do the following:
a. Sketch a graph of the equation
b> identify the vertex
compare the graph of y=f(x) to the graph of y=x^2 state any transformation used

Answer 11

it's just a parabola with a vertex of (0,-1), pointing upward. Transformation is a shift down along the y axis i guess, not sure what the terms are..

Answer 12

so when graphing what points so I use with the parabola.

Answer 13

transformations would be down, left, right or up?

Answer 14

Rather than using Calculus, we can use basic Pre-Calculus mathematics in order to find the maximum point. We know that the leading coefficient is negative, hence the vertex of the form must be a maximum. We can find this coordinate for the maximum by Completing-the-square and re-writing the function in vertex form.

\[-x^2+4x+9=-(x-2)^2+13\] This means that the coordinate of the vertex (maximum in this case) is (2, 13).

@PECKCE

Answer 15

transformation is down

Answer 16

ok let's generate a bunch of points for you to use:
(0,-1)
(1,0);(-1,0)
(2,3);(-2;3), etc.

Answer 17

Which means the maximum y-value is y = 13, and it occurs at x = 2.
Do you understand? I'm hoping you're familiar with completing the square/vertex form.

@PECKCE

Answer 18

thank you @genius12 it was very kind of you, I am not familiar with that but @bahrom7893 explain it to me and came to the same solution.

Answer 19

@PECKCE What @bahrom7893 has done is completely valid but it requires you to know Calculus. If you haven't covered Calculus, the approach would seem difficult to understand.

And you should be familiar with completing the square/vertex form, if you're not, go here to get a tutorial:

http://www.mathsisfun.com/algebra/completing-square.html

That will teach you exactly what to do and what completing the square means.