Question

1. no calculator at what point on the graph of $y = \frac{1}{2}x^{2}$ is the tangent line parallel to the line $2x - 4y = 3$? (a) $(\frac{1}{2},-\frac{1}{2})$ (b) $(\frac{1}{2},\frac{1}{8})$ (c) $(1,-\frac{1}{4})$ (d) $(1,\frac{1}{2})$ (e) $(2,2)$

Explanation:

Step1: Find the slope of the given line

Rewrite the line $2x - 4y=3$ in slope - intercept form $y = mx + b$.
$4y=2x - 3$, so $y=\frac{1}{2}x-\frac{3}{4}$. The slope $m$ of this line is $\frac{1}{2}$.

Step2: Find the derivative of the function $y=\frac{1}{2}x^{2}$

Using the power rule $(x^n)'=nx^{n - 1}$, if $y=\frac{1}{2}x^{2}$, then $y'=x$.

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

Since the tangent line is parallel to the given line, their slopes are equal. Set $y' = x=\frac{1}{2}$.

Step4: Find the $y$-coordinate

Substitute $x = \frac{1}{2}$ into the function $y=\frac{1}{2}x^{2}$. Then $y=\frac{1}{2}\times(\frac{1}{2})^{2}=\frac{1}{8}$.

Answer:

B. $(\frac{1}{2},\frac{1}{8})$