Question

on a separate piece of paper, sketch the graph of the parabola y = x^2+7. on the same graph, plot the point (0, - 5). note that there are two tangent lines of y = x^2+7 that pass through the point (0, - 5). specifically, the tangent line of the parabola y = x^2+7 at the point (a, a^2 + 7) passes through the point (0, - 5) where a>0. the other tangent line that passes through the point (0, - 5) occurs at the point (-a, a^2 + 7). find the number a.

Explanation:

Step1: Find the derivative of the parabola

The derivative of $y = x^{2}+7$ with respect to $x$ is $y'=2x$. The slope of the tangent line at the point $(a,a^{2}+7)$ is $m = 2a$.

Step2: Use the point - slope form of a line

The point - slope form of a line is $y - y_1=m(x - x_1)$. We know the line passes through $(a,a^{2}+7)$ and $(0, - 5)$. So the slope $m=\frac{(a^{2}+7)-(-5)}{a - 0}=\frac{a^{2}+12}{a}$.

Step3: Set the two expressions for the slope equal

Since the slope of the tangent line is $m = 2a$ and also $m=\frac{a^{2}+12}{a}$, we set $2a=\frac{a^{2}+12}{a}$. Cross - multiply to get $2a^{2}=a^{2}+12$.

Step4: Solve the resulting equation

Subtract $a^{2}$ from both sides of the equation $2a^{2}=a^{2}+12$. We have $2a^{2}-a^{2}=12$, which simplifies to $a^{2}=12$. Since $a>0$, then $a = \sqrt{12}=2\sqrt{3}$.

Answer:

$2\sqrt{3}$