Use 4x-13 to find an equation of the line tangent to the curve at x=3?

Question

I found the derivative of the function y=2x^2- 13x +5 and the answer is 4x-13 I then have to use that answer to find an equation of the line tangent to the curve at x=3

if you're good at math please help me with this problem! THANK YOU!

Answer 1

f(x) = 2x^2-13x+5
f'(x) = 4x - 13

f'(3) = -1
f(3) = -16

You're looking for a line with slope m=-1 that passes thru (3,-16)
y -(-16) = -1*(x-3)
y = -x-13

Answer 2

The slope of a tangent line is the negative reciprocol of the slope of the function. The slope of the function at x=3 is 4*3-13 or -1. The negative reciprocol of this is 1. So you are looking for an equation with a slope of one that passes through the curve where x=3. y=-16 when x=3. Use y=mx+b to solve. m=1 so find b with x=3 and y=-16. -16=1x3 + b. b=-19. So the answer is...

y = x - 19.

Answer 3

y=2x^2-13x+5 , dy/dx= 4x-13. since dy/dx means the slope of a tangent to the curve at a certain point, now the certain point is 3. so the gradient of a tangent to the curve at x=3 is -1. when, x=3, y is -16. eoquation is y+x=13