Question

The tangent line to the vertex of p(x)=4x2+8x+1 is:

Answer 1

the vertex of a parabola is always a max or min value. In this case because the parabola points upwards it is a min. At max/min of p(x)
p'(x)=8x+8=0
x=-1
p(-1)=4-8+1=-3
for point-slope form:
y+3=p'(-1)(x+1)=(0)(x+1)=0
y=-3

Answer 2

the vertex occurs where dp/dx=0

Answer 3

this is dp(x)/dx=0=8x+8
so, x=-1

Answer 4

the corresponding y coordinate is p(-1)=4(-1)2+8(-1)+1=-3

Answer 5

the vetx is (-1,-3)

Answer 6

the tangent is a line with forumla y=mx+b. it passes through the point (-1,-3) and it's slope (m) is given by dp/dx at x=-1