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OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

really?

OpenStudy (anonymous):

u sure?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

@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)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

but there is no way to do this without calculus

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (dumbcow):

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

OpenStudy (anonymous):

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

OpenStudy (dumbcow):

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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