ap calculus bc mcq practice test b\n18. in the xy - plane, the line x + y = k, where k is a constant, is…

ap calculus bc mcq practice test b\n18. in the xy - plane, the line x + y = k, where k is a constant, is tangent to the graph of y = x² + 3x + 1. what is the value of k?\n(a) -3 (b) -2 (c) -1 (d) 0 (e) 1\no a o b o c o d o e

ap calculus bc mcq practice test b\n18. in the xy - plane, the line x + y = k, where k is a constant, is tangent to the graph of y = x² + 3x + 1. what is the value of k?\n(a) -3 (b) -2 (c) -1 (d) 0 (e) 1\no a o b o c o d o e

Answer

Explanation:

Step1: Rewrite the line equation

Rewrite $x + y=k$ as $y=-x + k$. The slope of this line is $m=-1$.

Step2: Find the derivative of the quadratic - function

Differentiate $y = x^{2}+3x + 1$ with respect to $x$. Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $y^\prime=2x + 3$.

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

Since the slope of the tangent line is $-1$, set $2x+3=-1$. Solve for $x$: [ \begin{align*} 2x+3&=-1\ 2x&=-4\ x&=-2 \end{align*} ]

Step4: Find the $y$ - value on the parabola at $x = - 2$

Substitute $x=-2$ into $y=x^{2}+3x + 1$. [ \begin{align*} y&=(-2)^{2}+3\times(-2)+1\ &=4-6 + 1\ &=-1 \end{align*} ]

Step5: Find the value of $k$

Since the line $y=-x + k$ is tangent to the parabola at the point $(-2,-1)$, substitute $x=-2$ and $y=-1$ into $y=-x + k$. [ \begin{align*} -1&=-(-2)+k\ -1&=2 + k\ k&=-3 \end{align*} ]

Answer:

A. -3