Find the equation of the tangent to the parabola y = x^2 - 5x - 3 that is parallel to the line 3x - y - 7 = 0.

Question

Working out please!

Step by step..

Thank you!

Answer 1

y=x^2-5x-3

the slope of the tangent line at a point of the parabola is the derivative of y=x^2-5x-3

dy/dx=2x-5

parallel lines have the same slope

3x-y-7=0 becomes y=3x-7 whose slope is 3

so 3=2x-5 and 2x=8 and x=4

y=4^2-5*4-3

y=16-23=-7

the tangent to the parabola is at (4,-7)

the tangent has equation y=3x+b

using (4,-7)...

-7=3(4)+b

-7=12+b

b=-19

the equation of the tangent parallel to the given line is y=3x-19

Answer 2

Hello Elena,

Here are the steps I am going to take:

i) Put the linear equation in y = mx + b form, so that I can read off the slope value.

ii) Find the derivative of the quadratic, set that equal to the slope value, and solve for the x-coordinate

iii) Use the x coordinate value in the quadratic to find the corresponding y-coordinate

iv) Use the (x,y) coordinate with the slope value to write the tangent line equation

i) 3x - y - 7 = 0

→ y = 3x - 7

→ m = 3

ii) y = x² - 5x - 3

→ y' = 2x - 5

Setting 2x - 5 = 3

→ 2x = 8

→ x = 4

iii) For x = 4

→ y = (4)² - 5(4) - 3

→ y = 16 - 20 - 3

→ y = -7

→ (x,y) = (4 , -7)

iv) Using point-slope form, we get

y + 7 = 3(x - 4) for the tangent line to the curve.

Thank you for the question, and have a great evening!

~Michael Ehlers