Question

a. find the derivative function f’ for the function f. b. find an equation of the line tangent to the graph of f at (a,f(a)) for the given value of a. f(x)=2x^2 - 7x + 5, a = 2 a. f’(x)= b. y=

Explanation:

Step1: Apply power - rule for derivative

The power - rule states that if $y = x^n$, then $y^\prime=nx^{n - 1}$. For $f(x)=2x^{2}-7x + 5$, we have:
$f^\prime(x)=\frac{d}{dx}(2x^{2})-\frac{d}{dx}(7x)+\frac{d}{dx}(5)$.
Since $\frac{d}{dx}(2x^{2})=2\times2x^{2 - 1}=4x$, $\frac{d}{dx}(7x)=7$, and $\frac{d}{dx}(5)=0$, then $f^\prime(x)=4x - 7$.

Step2: Find the slope of the tangent line at $x = a$

We are given $a = 2$. Substitute $x = 2$ into $f^\prime(x)$ to find the slope $m$ of the tangent line.
$m=f^\prime(2)=4\times2-7=8 - 7=1$.

Step3: Find the point on the curve

Substitute $x = 2$ into $f(x)$ to find the $y$ - coordinate of the point of tangency.
$f(2)=2\times2^{2}-7\times2 + 5=2\times4-14 + 5=8-14 + 5=-1$. So the point $(a,f(a))=(2,-1)$.

Step4: Use the point - slope form of a line

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)$ is a point on the line and $m$ is the slope.
Here, $x_1 = 2$, $y_1=-1$, and $m = 1$.
$y-(-1)=1\times(x - 2)$.
Simplify to get $y+1=x - 2$, or $y=x-3$.

Answer:

a. $4x - 7$
b. $y=x - 3$