Question

Find the equations of all lines tangent to y = 9–x^2 that pass through the point (1, 12).

Answer 1

gradient of tangent = dy/dx = - 2x
= -2 at the point (1,12)
using
y-y1 = m(x-x1)
y - 12 = -2(x-1)
y + 2x = 14 is the equation of the tangent

Answer 2

we didn't learn dy/dx.. :(

Answer 3

really?

Answer 4

u sure?

Answer 5

then how would you know the slope of a tangent line, or even what "tangent line" means?

Answer 6

@jimmyrep i don't think this is what the question asks for. it asks for lines tangent to the graph that pass through the point (1,12)

Answer 7

also notice that (1,12) is not on the graph. the slope of a line tangent to the graph of
\[y=9-x^2\] at a point
\[(x_1, y_1)\]
\[-2x_1\]

Answer 8

so you need to solve
\[-2x_1=\frac{12-(9-x_1^2)}{1-x_1}\] for for
\[x_1\]

Answer 9

but there is no way to do this without calculus

Answer 10

you get
\[2x_1(x_1-1)=3+x_1^2\]
\[x_1^2+2x_1-3=0\]
\[x_1=-1,x_1=3\]

Answer 11

Where did you get -2x1?.. and how did you set up your equation?

Answer 12

From taking the derivative of the function: 9-x^2
what class is this for?

Answer 13

Calc lol.. the organization of our book is confusing..

Answer 14

oh ok
yeah so the slope at any point is -2x
then follow satellites steps to find the points where the tangent line goes through (1,12)