How do you find an equation at a tangent line to the curve #y = (5+4x)^2# at point p = (7, 4)?

y1=?
y2=?

ahh pleasee help...i am soo confused here :/?

Question

How do you find an equation at a tangent line to the curve #y = (5+4x)^2# at point p = (7, 4)?

y1=?
y2=?

ahh pleasee help...i am soo confused here :/?

Christopher Agresta · Accepted Answer

To find the equation of the tangent line to the curve #y = (5 + 4x)^2# at point #(7,4)# we need to find the gradient/slope of the tangent line at the point #(7,4)# to do this, we need to differentiate our function. Note that the point does not exist on that curve, so I will use a point that does exist on the curve, which would be #(7,1089)#
I got to that point by simply setting #x=7# in the equation to get #y# #dy/dx f(x) = dy/dx (5 + 4x)^2# #f'(x) = 8(5+4x)# then we sub in the #x# value from the given point. #f'(7) = 8(5+4(7)) = 264# now we have got the gradient or #m# at the point where #x=7# 
we can start using our formula for a straight line, which is: #y = mx + c# and we need to solve for #c# so we use our known point. #1089 = (264)(7) + c#
#c = - 759# thus the equation to the line tangent to the curve at the point where #x=7# would be. #y = 264x - 759#

Isabella Ackerman · Answer

undefined To find the equation of the tangent line to the curve y = (5+4x)^2 at point p = (7, 4), we need to find the slope of the tangent line and the coordinates of the point of tangency.

First, we find the derivative of the function y = (5+4x)^2 with respect to x.

dy/dx = 2(5+4x)(4) = 8(5+4x)

Next, we substitute the x-coordinate of the point of tangency, which is 7, into the derivative to find the slope of the tangent line at that point.

dy/dx = 8(5+4(7)) = 8(5+28) = 8(33) = 264

So, the slope of the tangent line at point p is 264.

Now, we can use the point-slope form of a linear equation to find the equation of the tangent line.

y - y1 = m(x - x1)

Substituting the coordinates of point p and the slope we found, we have:

y - 4 = 264(x - 7)

Expanding and simplifying:

y - 4 = 264x - 1848

y = 264x - 1844

Therefore, the equation of the tangent line to the curve y = (5+4x)^2 at point p = (7, 4) is y = 264x - 1844.

How do you find an equation at a tangent line to the curve #y = (5+4x)^2# at point p = (7, 4)? y1=? y2=? ahh pleasee help...i am soo confused here :/?