Question

Find equations of both lines through the point (2,-3) that are tangent to the parabola y+x^2+x.

Help would be much appreciated :)

Answer 1

oops y=x^2+x

Answer 2

i hope to god you can use calculus

Answer 3

you can do it without, but it is a real pain

Answer 4

yeah, no prob!

Answer 5

ok whew

Answer 6

so the derivative is \(2x+1\) and if the point is tangent to the graph, that means that the slope has to agree with \(2x+1\) for whatever \(x\) value you have
the slope through a point on the graph \(y=x^2+x\) looks like \((x,x^2+x)\) and the slope between that point and \((2,-3)\) will be
\[\frac{x^2+x+3}{x-2}\]

Answer 7

ok i said that wrong, but you get the idea i hope
in any case, your job is to solve
\[\frac{x^2+x+3}{x-2}=2x+1\] for \(x\)
you can see that you will get a quadratic equation, which is why you have two solutions

Answer 8

thank you so much!

Answer 9

yw
sorry my explanation was so convoluted, hope it makes sense

Answer 10

it did :)

Answer 11

Here is solution not using Calculus
Let g(x)= -3 + m(x-2) be any line going through the point (2,-3)
Let us find the points of intersections of this line with the parabola. We have to
slove
\[
x^2 + x = -3 + m(x-2)\\
x^2 +x(1-m)+3+2m==0\\

\]
to be tangent, the two roots have to be equal. Hence the discriminant
should be zero
\[
(1-m)^2 -4(3+2m)==0\\
m=-1\\
m=11

\]
This will give you two tangents
\[
y1=-1-x\\
y2=-3+11 (-2+x)\\
\]

Answer 12

@satellite73