Let f(x)=7x^2-4x+6. Then the quotient (f(5+h)-f(5))/h can be simplified to ah+b for a=? and b=?

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Let f(x)=7x^2-4x+6. Then the quotient (f(5+h)-f(5))/h can be simplified to ah+b for a=? and b=?

Mathematics
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are we looking for the equation of the tangent line at f(5)?
I'm not sure, those are the only instructions I was given and I was never shown any sample problems, but I think that would make sense
have you learned to do derivatives yet? or is that what you getting up to?

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Other answers:

ive learned
good, the slope of the line is the derivative of the function at f'(5).
as far as I know, "a" would equal f'(5)
I dont think he is working on a derivative otherwise he would be taking the limit as h approaches 0
b might = f(5) - f'(5)(h)... nadeems got a valid point :)
i understand how to find the derivative, i'm just confused what the ah+b is representing in all of this
f(5 + h) - f(5) = 7(5 + h)^2 - 4(5 + h) + 6 - 7(25) + 4(5) - 6 = 7(25 + 10h + h^2) - 20 - 4h - 155 = 175 + 70h +7h^2 -175 -4h 7h^2 + 66h [f(5 + h) - f(5)] = (7h^2 + 66h)/h = 7h +66
might be a dummy form of the line equation... maybe h needs to be kept in as the variable?
\[f(x)=7x^2-4x+6\] \[f(5+h)=7(5+h)^2-4(5+h)+6\] \[f(5)=7(5^2)-4(5)+6\]
now plug in these values in to the function: \[\frac{f(5+h)-f(5)}{h}\rightarrow \frac{7(h^2+10h+25)-20-4h+6-159}{h}\]
i see now
now simplify: \[\frac{7h^2+66h}{h}\rightarrow \frac{7h^2}{h}+\frac{66h}{h}\rightarrow 7h+66\]
so a=7 and b=66
great, thanks!
no problem

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