How do you find the equation of a line tangent to the function #y=x^2+3x-1# at x=0?

Question

Elijah Abram · Accepted Answer

#y=3x-3# The equation of a straight line with given gradient #m# and one known point #(x_1,y_1)# is: #y-y_1=m(x-x_1)# 1) for #y=x^2+3x-1 # at #x=0# #y(0)=0+0-1=-1# #(x_1,y_1)=(0,-1)# 2) the gradient of the tangent at the given point is found by differentiating. #y=x^2+3x-1 # #(dy)/(dx)=2x+3# #((dy)/(dx))_(x=0)=2xx0+3=3# 3) use #y-y_1=m(x-x_1)# #y-(-1)=3(x-0)# #y+1=3x# #:.y=3x-1#

Elijah Abram · Answer

undefined To find the equation of a line tangent to a function at a specific point, you need to find the derivative of the function and evaluate it at the given x-value.

The derivative of the function y = x^2 + 3x - 1 is y' = 2x + 3.

To find the slope of the tangent line at x = 0, substitute x = 0 into the derivative: y'(0) = 2(0) + 3 = 3.

Since the slope of the tangent line is 3, and it passes through the point (0, f(0)), which is (0, -1) in this case, you can use the point-slope form of a linear equation to find the equation of the tangent line.

Using the point-slope form, the equation of the tangent line is y - y1 = m(x - x1), where m is the slope and (x1, y1) is the given point.

Plugging in the values, the equation of the tangent line is y - (-1) = 3(x - 0), which simplifies to y + 1 = 3x.