How do you find the equation of tangent line to the curve #y=2x^2+4x-3# at the point (1,3)?

Question

Isaiah Allen · Accepted Answer

# y=8x-5#

The gradient of the tangent to a curve at any particular point is give by the derivative of the curve at that point.

so If #y=2x^2+4x-3# then differentiating wrt #x# gives us:

#dy/dx = 4x+4#

When #x=1 => y=2+4-3=3# (so #(1,3)# lies on the curve)
and #dy/dx=4+4=8#

So the tangent we seek passes through #(1,3)# ad has gradient 8 so using #y-y_1=m(x-x_1)# the equation we seek is;

# y-3=8(x-1) #
# :. y-3=8x-8#
# :. y=8x-5#

We can confirm this graphically:

Christopher Agresta · Answer

undefined To find the equation of the tangent line to the curve y=2x^2+4x-3 at the point (1,3), we need to find the slope of the tangent line at that point.

To find the slope, we take the derivative of the given function y=2x^2+4x-3 with respect to x.

The derivative of y=2x^2+4x-3 is dy/dx = 4x+4.

Substituting x=1 into the derivative, we get dy/dx = 4(1)+4 = 8.

So, the slope of the tangent line at the point (1,3) is 8.

Using 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, we can substitute the values to find the equation of the tangent line.

Substituting x1=1, y1=3, and m=8 into the equation, we get y-3 = 8(x-1).

Simplifying the equation, we have y-3 = 8x-8.

Rearranging the equation, we get y = 8x-5.

Therefore, the equation of the tangent line to the curve y=2x^2+4x-3 at the point (1,3) is y = 8x-5.