Equation of the Tangent Line

[nycmathguy]

Find the equation of the tangent line to the curve
f(x) = 2x^2 + 5x - 3 at x = 2.

P. S. Make up a similar problem for me to do on my own.

Thank you.

 

[MathLover1]

Find the equation of the tangent line to the curve

f(x) = 2x^2 + 5x - 3 at x = 2.

derivative of f(x) will be a slope of the tangent line

f'(x) = 2*2x^1 + 1*5x^0 - 0*3
f'(x) = 4x + 5

at x = 2, the slope of the tangent line is f '(2)

f'(2) = 4*2 + 5
f'(2) = 13 -> a slope of tangent line


You know that the tangent line shares at least one point with the original equation, f(x) = 2x^2 + 5x - 3. Since the line you are looking for is tangent to f(x) = 2x^2 + 5x - 3 at x = 2, you know the x coordinate for one of the points on the tangent line. By plugging the x coordinate of the
shared point into the original equation you have:
f(2) = 2*2^2 + 5*2 - 3 =15

Therefore, you have found the coordinates, (2, 15), for the point shared by f(x) and the line
tangent to f(x) at x = 2. Now you have a point on the tangent line and the slope of the
tangent line.

now use slope point formula ti find equation af tangent

y-15=13(x-2)
y=13(x-2)+15
y=13x-26+15
y=13x-11

for you: find the tangent line to the curve y=x^2-9x at the point where x=1

 

[nycmathguy]

Find the equation of the tangent line to the curve

f(x) = 2x^2 + 5x - 3 at x = 2.

derivative of f(x) will be a slope of the tangent line

f'(x) = 2*2x^1 + 1*5x^0 - 0*3
f'(x) = 4x + 5

at x = 2, the slope of the tangent line is f '(2)

f'(2) = 4*2 + 5
f'(2) = 13 -> a slope of tangent line


You know that the tangent line shares at least one point with the original equation, f(x) = 2x^2 + 5x - 3. Since the line you are looking for is tangent to f(x) = 2x^2 + 5x - 3 at x = 2, you know the x coordinate for one of the points on the tangent line. By plugging the x coordinate of the
shared point into the original equation you have:
f(2) = 2*2^2 + 5*2 - 3 =15

Therefore, you have found the coordinates, (2, 15), for the point shared by f(x) and the line
tangent to f(x) at x = 2. Now you have a point on the tangent line and the slope of the
tangent line.

now use slope point formula ti find equation af tangent

y-15=13(x-2)
y=13(x-2)+15
y=13x-26+15
y=13x-11

for you: find the tangent line to the curve y=x^2-9x at the point where x=1

I will answer "for you" later today. On the Coney Island boardwalk now.

 

[nycmathguy]

Find the equation of the tangent line to the curve

f(x) = 2x^2 + 5x - 3 at x = 2.

derivative of f(x) will be a slope of the tangent line

f'(x) = 2*2x^1 + 1*5x^0 - 0*3
f'(x) = 4x + 5

at x = 2, the slope of the tangent line is f '(2)

f'(2) = 4*2 + 5
f'(2) = 13 -> a slope of tangent line


You know that the tangent line shares at least one point with the original equation, f(x) = 2x^2 + 5x - 3. Since the line you are looking for is tangent to f(x) = 2x^2 + 5x - 3 at x = 2, you know the x coordinate for one of the points on the tangent line. By plugging the x coordinate of the
shared point into the original equation you have:
f(2) = 2*2^2 + 5*2 - 3 =15

Therefore, you have found the coordinates, (2, 15), for the point shared by f(x) and the line
tangent to f(x) at x = 2. Now you have a point on the tangent line and the slope of the
tangent line.

now use slope point formula ti find equation af tangent

y-15=13(x-2)
y=13(x-2)+15
y=13x-26+15
y=13x-11

for you: find the tangent line to the curve y=x^2-9x at the point where x=1

I gotta find y'.

So, y' = 2x - 9.

At x = 1, the slope of the tangent line is f '(1).
So, f '(1) = 2(1) - 9 or -7.

I now need to find the y-coordinate of the point we need to plug into the point-slope formula.

In the given function, let x be 1.

y = (1)^2 - 9(1)

y = 1 - 9 or -8.

Our point is (1, -8).

Plug into point-slope formula to find the equation of the tangent line to the curve.

y - y_1 = m(x - x_1)

y - (-8) = -7(x - 1)

y + 8 = -7x + 7

y = -7x + 7 - 8

y = -7x - 1

You say?