How do you find the equations to the two tangent lines to the graph of #f(x)=5x^2# that pass through the point (-1,-1)?

Question

Elijah Abram · Accepted Answer

#y = (5+2sqrt(30))x - 5(1/2+sqrt(30)/5)^2#
and
#y = (5-2sqrt(30))x - 5(1/2-sqrt(30)/5)^2# A tangent line of #y = f(x)# at some point #(x_0, f(x_0))# on the curve will be a line with a slope of #f'(x_0)#. Plugging this into the point-slope formula of a line, we get #y - f(x_0) = f'(x_0)(x - x_0)# #=> y = f'(x_0) - x_0f'(x_0) + f(x_0)# Given #f(x) = 5x^2#, we get #f'(x) = 10x#, meaning a tangent line at #(x, y) = (x_0, 5x_0^2)# will be a line of the form #y = (10x_0)x - x_0(10x_0) + 5x_0^2# #=> y = 10x_0x - 5x_0^2# We wish to find any such lines which pass through the point #(-1, -1)#. Substituting in #(x, y) = (-1, -1)# gives us #(-1) = 10x_0(-1) - 5x_0^2# #=> 5x_0^2 + 10x_0 - 1 = 0# #=> x_0 = (-5+-sqrt(10^2-4(-1)(5)))/(2(5))# #=(-5+-2sqrt(30))/10# #=1/2+-sqrt(30)/5# Thus, substituting those values back into our equation for a tangent line #y = 10x_0x - 5x_0^2#, we get our two tangent lines as #y = (5+2sqrt(30))x - 5(1/2+sqrt(30)/5)^2# and #y = (5-2sqrt(30))x - 5(1/2-sqrt(30)/5)^2#

Christopher Allers · Answer

undefined To find the equations of the two tangent lines to the graph of f(x) = 5x^2 that pass through the point (-1, -1), we can follow these steps:

1. Find the derivative of f(x) with respect to x, denoted as f'(x).
2. Substitute the x-coordinate of the given point (-1, -1) into f'(x) to find the slope of the tangent line at that point.
3. Use the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, to write the equation of the tangent line passing through (-1, -1).
4. Repeat steps 2 and 3 to find the equation of the second tangent line.

Let's calculate the equations of the two tangent lines:

1. Differentiating f(x) = 5x^2 with respect to x, we get f'(x) = 10x.

2. Substituting x = -1 into f'(x), we find f'(-1) = 10(-1) = -10. Therefore, the slope of the tangent line at (-1, -1) is -10.

3. Using the point-slope form, we have y - (-1) = -10(x - (-1)), which simplifies to y + 1 = -10(x + 1). This is the equation of the first tangent line.

4. To find the equation of the second tangent line, we repeat steps 2 and 3 for a different point on the graph. Let's choose x = 1.

Substituting x = 1 into f'(x), we find f'(1) = 10(1) = 10. Therefore, the slope of the tangent line at (1, 5) is 10.

Using the point-slope form, we have y - (-1) = 10(x - (-1)), which simplifies to y + 1 = 10(x + 1). This is the equation of the second tangent line.

Hence, the equations of the two tangent lines to the graph of f(x) = 5x^2 that pass through the point (-1, -1) are:
1. y + 1 = -10(x + 1)
2. y + 1 = 10(x + 1)