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

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

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

really?

u sure?

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

@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)

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 \[(x_1, y_1)\] \[-2x_1\]

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

but there is no way to do this without calculus

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

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

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

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

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)

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