How do you find the equations of the tangent to the curve #y=x^2 + 2x -3# pass through the point (0,-7)?

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Answer 1 · 2017-03-09T16:22:33

The two equations are

#y = 6x - 7#
#y = -2x - 7#

The derivative is

#y = 2x + 2#

The slope will be given by #m= 2(a) + 2#. The point will be #(a, a^2 + 2a - 3)#.

#y - y_1 = m(x- x_1)#

#y - (a^2 + 2a - 3) = (2a + 2)(x - a)#

We can replace #(x, y)# with our point #(0, -7)#.

#-7 - a^2 -2a + 3 = (2a + 2)(0 - a)#

#-7 - a^2 - 2a + 3 = -2a^2 - 2a#

#a^2 - 4 = 0#

#(a + 2)(a - 2) = 0#

#a = 2 and -2#

Now use this to compute the actual equation.

For when #a = 2#

We have #m = 2(2) + 2 = 6# and the point will be #(2, 5)#.

Therefore, the equation is

#y - y_1 = m(x - x_1)#

#y - 5 = 6(x - 2)#

#y - 5 = 6x - 12#

#y = 6x - 7#

For when #a = -2#

We have #m = 2(-2) + 2 = -2# and the point will be #(-2, -3)#.

Therefore, the equation is

#y - y_1 = m(x- x_1)#

#y - (-3) = -2(x - (-2))#

#y + 3 = -2(x + 2)#

#y + 3 = -2x - 4#

#y = -2x - 7#

Hopefully this helps!