At what point on the graph of #y = (1/2)x^2 - (3/2)# is the tangent line parallel to the line 4x - 8y = 5?

Question

Isaiah Allen · Accepted Answer

#(1/2, -11/8)# The slope of the given line is #1/2#. The slope of the tangent to the given curve #y=(1/2)x^2-(3/2)#, at (x, y) on the curve, is #y'=(1/2)(2x)=x. So, for the parallel tangent the slope #= x = 1/2#. The corresponding #y= (1/2)(1/2)^2-(3/2)=(1/8)-(3/2)=-(11/8)#. Answer: #(1/2, -11/8)#. .

Christopher Allers · Answer

undefined To find the point on the graph of y = (1/2)x^2 - (3/2) where the tangent line is parallel to the line 4x - 8y = 5, we need to determine the slope of the tangent line and compare it to the slope of the given line.

The slope of the given line can be found by rearranging it into slope-intercept form (y = mx + b), where m represents the slope.

4x - 8y = 5
-8y = -4x + 5
y = (1/2)x - (5/8)

Comparing this equation to y = mx + b, we can see that the slope of the given line is 1/2.

To find the slope of the tangent line to the graph of y = (1/2)x^2 - (3/2), we need to take the derivative of the equation with respect to x.

dy/dx = x

Setting this derivative equal to the slope of the given line (1/2), we have:

x = 1/2

Substituting this value of x back into the original equation, we can find the corresponding y-coordinate:

y = (1/2)(1/2)^2 - (3/2)
y = 1/8 - (3/2)
y = -23/8

Therefore, the point on the graph of y = (1/2)x^2 - (3/2) where the tangent line is parallel to the line 4x - 8y = 5 is (1/2, -23/8).