Question

if (f(x)=4x^{2}-4x + 4), find (f(4)). use this to find the equation of the tangent line to the parabola (y = 4x^{2}-4x + 4) at the point ((4,52)). the equation of this tangent line can be written in the form (y=mx + b) where (m) is: and where (b) is:

Explanation:

Step1: Find the derivative of $f(x)$

Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, if $f(x)=4x^{2}-4x + 4$, then $f'(x)=\frac{d}{dx}(4x^{2})-\frac{d}{dx}(4x)+\frac{d}{dx}(4)$.
$f'(x)=4\times2x-4\times1 + 0=8x-4$.

Step2: Evaluate $f'(4)$

Substitute $x = 4$ into $f'(x)$.
$f'(4)=8\times4-4=32 - 4=28$.

Step3: Find the slope $m$ of the tangent line

The slope $m$ of the tangent line to the curve $y = f(x)$ at a point is equal to the value of the derivative at that point. So $m=f'(4)=28$.

Step4: Find the value of $b$

The equation of the line is $y=mx + b$, and the line passes through the point $(4,52)$. Substitute $x = 4$, $y = 52$, and $m = 28$ into $y=mx + b$.
$52=28\times4 + b$.
$52=112 + b$.
Solve for $b$: $b=52 - 112=-60$.

Answer:

$f'(4)=28$
$m = 28$
$b=-60$