finding an equation of a tangent line. in exercises 37 - 42, find an equation of the line that is tangent to…

Explanation:

Step1: Rewrite the line equation in slope - intercept form

The given line is $x + y=0$, which can be rewritten as $y=-x$. The slope of this line $m=-1$.

Step2: Differentiate the function

The function $f(x)=-\frac{1}{4}x^{2}$. Using the power - rule for differentiation, if $y = ax^{n}$, then $y^\prime=anx^{n - 1}$. So $f^\prime(x)=-\frac{1}{4}\times2x=-\frac{1}{2}x$.

Step3: Set the derivative equal to the slope of the given line

We want to find the $x$ - value where the slope of the tangent line to $y = f(x)$ is equal to the slope of the line $x + y = 0$. Set $f^\prime(x)=-1$. So $-\frac{1}{2}x=-1$, which gives $x = 2$.

Step4: Find the $y$ - value on the function

Substitute $x = 2$ into $f(x)=-\frac{1}{4}x^{2}$. Then $f(2)=-\frac{1}{4}\times2^{2}=-1$.

Step5: Use the point - slope form to find the tangent line equation

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 $m=-1$. So $y+1=-(x - 2)$, which simplifies to $y=-x + 1$.

Answer:

$y=-x + 1$