finding an equation of a tangent line in exercises 29 - 36, (a) find an equation of the tangent line to the…

Let's solve problem 29: $f(x)=x^{2}+3$ at the point $(-1,4)$.

Explanation:

Step1: Find the derivative of the function

The derivative of $f(x)=x^{2}+3$ using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$ is $f^\prime(x)=2x$.

Step2: Evaluate the derivative at the given $x$ - value

Substitute $x=-1$ into $f^\prime(x)$. So $f^\prime(-1)=2\times(-1)=-2$. This is the slope $m$ of the tangent line.

Step3: Use the point - slope form of a line

The point - slope form is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(-1,4)$ and $m = - 2$. Substitute these values: $y - 4=-2(x+1)$.

Step4: Simplify the equation

Expand the right - hand side: $y - 4=-2x-2$. Add 4 to both sides to get the equation of the tangent line in slope - intercept form: $y=-2x + 2$.

Answer:

The equation of the tangent line is $y=-2x + 2$.

For part (b), to graph the function $y=x^{2}+3$ and its tangent line $y=-2x + 2$ using a graphing utility (such as a graphing calculator or software like Desmos), you would enter the two functions separately. The graph of $y=x^{2}+3$ is a parabola opening upwards with vertex at $(0,3)$, and the graph of $y=-2x + 2$ is a straight line with slope $-2$ and $y$ - intercept $2$. They should intersect at the point $(-1,4)$.

For part (c), most graphing utilities have a "tangent" feature. You would input the function $y=x^{2}+3$ and the point $x=-1$. The utility should then display the tangent line $y=-2x + 2$, confirming the result from part (a).