# To get an idea of how fonts are defined (Calculus)....?

The function should be in the form f(x)=c(x-2.5)^2+k. Use f(1) =1=f(4), f'(1)=1, f'(4)=-1 to solve for constants c and k. Please show work/explain thank you

Update:

It does not have a derivative at x=1 or x=4

### 1 Answer

Relevance

- Jeff AaronLv 71 month ago
f(x) = c(x^2 - 5x + 6.25) + k

f(x) = cx^2 - 5cx + 6.25c + k

f'(x) = 2cx - 5c

f'(1) = 2c*1 - 5c = 1

2c - 5c = 1

-3c = 1

c = -1/3

f(x) = (-1/3)(x - 2.5)^2 + k

f(1) = (-1/3)(1 - 2.5)^2 + k = 1

(-1/3)(-1.5)^2 + k = 1

(-1/3)2.25 + k = 1

-0.75 + k = 1

k = 1 + 0.75

k = 1.75

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