Equation ofthe tangent

[BonBon101]

Homework Statement

Determine the equation of the tangent line and the equation of the normal line to the curve y at the point (-2,-5)

y=1+x+x^2

The Attempt at a Solution

y=1+x-x^2 point (-2,-5)

y=1+x-x^2
y=1-2x

sub in -2 for x
y=1-2(-2)y=1+4
y=5

then i use the formula y-y1=m(x-x1) to find the tangent
y+5=5(x+20
y+5=5x+10
y=5x+5

5x-y+5=0 (equation of the tangent)

im wondering is this right?
and how do i go about finding the equation of the normal line?

 

[Mark44]

Homework Statement

Determine the equation of the tangent line and the equation of the normal line to the curve y at the point (-2,-5)

y=1+x+x^2

The Attempt at a Solution

y=1+x-x^2 point (-2,-5)

y=1+x-x^2
y=1-2x

This should be y' = 1 - 2x or dy/dx = 1 - 2x

sub in -2 for x
y=1-2(-2)y=1+4
y=5

This is the value of the derivative at x = -2.

then i use the formula y-y1=m(x-x1) to find the tangent
y+5=5(x+20

Typo above. You hit 0 instead of ).

y+5=5x+10
y=5x+5

5x-y+5=0 (equation of the tangent)

Either equation above is correct.
To check, is (-2, -5) a point on the line? Is the slope of the line 5? If the answer is yes to both questions, you have the right tangent line.

im wondering is this right?
and how do i go about finding the equation of the normal line?

What will be the slope of the normal line? This line must also go through the point (-2, -5). If you have the slope and a point on the line, you can do as you did before to find its equation.

Your computations were correct, but you didn't distinguish between y in your original curve and y' (or dy/dx), which is the derivative function. It's very important that you understand the difference between a function and its derivative, and that your work shows that you know the difference.