Find the equation of the line tangent to y = 2x2 – x + 4 at x = 3.
a. y = 11x – 14
b. y = 4x + 14
c. y = 4x + 3
d. y = 11x + 3

Y=2x²-x+4

1) we have to find the slope of this line.
we have to calculate the first derivative of this function.
y´=4x-1

The slope in x=3 will be: m(3)=4(3)-1=12-1=11.

2)We have to find a point of this line:
if x=3; ⇒y=2(3)²-3+4=18-3+4=19; Our point will be (3,19)

2) point-slope form: we need a point (x₀,y₀) and the slope.
y-y₀=m(x-x₀)
We know the slope (m=11) and the point (3,19); therefore:
y-19=11(x-3)
y=11x-33+19
y=11x-14

A. y=11x-14