Question

find the value of b in:
f(x)=-x^2+bx-75 when the max=25

Answer 1

max?

Answer 2

maximum value is at 25

Answer 3

try b = 20

Answer 4

how did you get that?

Answer 5

well actually you can use b = 20 or b = -20

Answer 6

can you show the math?

Answer 7

maximum value is at the vertex. the first coordinate of the vertex is always
\[-\frac{b}{2a}\] in this case since
\[a=-1\] the first coordinate of the vertex is
\[\frac{b}{2}\] then i replaced x by
\[\frac{b}{2}\] in the equation, set the result = 25, and solved for b. would you like the steps?

Answer 8

yes please

Answer 9

ok no problem, but i hope it is clear why i am replacing x by
\[\frac{b}{2}\]

Answer 10

can you explain why you are replacing it?

Answer 11

\[-(\frac{b}{2})^2+b\times \frac{b}{2}-75=25\]
\[-(\frac{b}{2})^2+b\times \frac{b}{2}=100\]
\[-\frac{b^2}{4}+\frac{b^2}{2}=100\]
\[\frac{b^2}{4}=100\]
\[b^2=400\]
\[b=\pm\sqrt{400}\]
\[b=\pm20\]

Answer 12

thanks so much!

Answer 13

if you have a quadratic that looks like
\[y=ax^2+bx+c\] then the maximum or minimum is always at the vertex (that is what it mean) and the first coordinate of the vertex is always
\[x=-\frac{b}{2a}\] so i know that the maximum occurs when
\[x=\frac{b}{2}\] in your example since a = -1. then i plug it in, and set it equal to 25 because you are told that the maximum is 25