a. find the slope of the tangent line to the graph of f(x)=3x - 3 at the point (1,0).\nb. find the slope…

Explanation:

Step1: Recall derivative formula

The derivative of a function $y = f(x)$ gives the slope of the tangent line. For $f(x)=3x - 3$, using the power - rule $\frac{d}{dx}(ax^n)=nax^{n - 1}$, where for $y = 3x-3$, the derivative $f^\prime(x)=\frac{d}{dx}(3x)-\frac{d}{dx}(3)$. Since $\frac{d}{dx}(3x)=3$ and $\frac{d}{dx}(3) = 0$, then $f^\prime(x)=3$.

Step2: Evaluate derivative at given point

The slope of the tangent line at any point on the graph of $y = f(x)$ is given by the value of the derivative at that point. Since $f^\prime(x)=3$ is a constant function, at the point $(1,0)$, the slope of the tangent line $m_{\tan}=f^\prime(1) = 3$.

Step3: Find slope - intercept equation

The slope - intercept form of a line is $y=mx + b$, where $m$ is the slope and $b$ is the y - intercept. We know $m = 3$ and the line passes through the point $(1,0)$. Substitute $x = 1$, $y = 0$ and $m = 3$ into $y=mx + b$. We get $0=3\times1 + b$, which simplifies to $b=-3$. So the equation of the tangent line is $y = 3x-3$.

Answer:

a. $m_{\tan}=3$ b. $y = 3x-3$