We have to find the equations of both lines through the point (2,-3) that are tangent to the parabola `y=x^2+x`
First we will find the derivative of y to find the slope m.
Now we find two points on the two tangent lines to y using: `(x,x^2+x)` and (2,-3). This is to show that (x,f(x)) is a point on the original equation f(x).
Now we have to find the slope of these two points and equate it to the derivative or slope of the original equation.
i.e. `x=5 , -1` ``
So now at x=5 the slope is m= 2(5)+1=11.So, the equation of the first tangent line through point (2,-3) is y+3=11(x-2)
at x=-1, the slope is m=2(-1)+1=-1 .So, the equation of the second tangent line through point (2,-3) is: y+3= -1(x-2).
So the equation of the two tangent lines to the parabola `y=x^2+x` passing through the point (2,-3) are:
`y=11x-25` and `y=-x-1`
The slope of the line tangent...
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