Calculus problem for my daughter!

Question

Identify an equation of the line tangent to y=2x^2+3 parallel to the 8x-y+3=0

Answer 1

First notice that the line 8x-y + 3= 0 can be written in slope-intercept form as y= 8x + 3, so has slope 8. Therefore every line parallel to it will also have slope 8. So you have to find the point on the curve y= 2x^2 +3 where the slope is 8. The derivative is the slope, so y' = 4x = 8, so x=2, and plugging in to the curve gives y=11. Thus the tangent line is the line with slope 8 passing through the point (2,11). Plugging the point into y=8x+b gives us b=-5, so the equation is y= 8x -5.