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I have a problem with derivatives, I've been trying to solve but I was not able to do it.

A parabola is tangent to the line $3x-y+6 = 0$ in the point $(0,6)$ and goes through the point $(1,0)$. Find the equation of the parabola supposing the equation is of the form $y = Ax^2 +Bx +C$ where $A B C$ are unknown.

I would like you to help me to solve it. So far I know $(1,0)$ is the vertex of the parabola and that $(0,6)$ is a point of the parabola,thus the directive is $x=0$

Any help would be appreciated.

Victor Castillo Torres
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  • What's the gradient of the line? What's an equation for the gradient of the parabola? – Jack Oct 24 '14 at 03:07
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    Find the slope of the tangent line. Then it follows that $y'(0)$ equals that slope. You also have $y(0) = 6$ and $y(1) = 0$ -> 3 equations, 3 unknowns. – Dylan Oct 24 '14 at 03:10

2 Answers2

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If you already have a tangent line, then taking the derivative of your given parabola form will yield

$$y'=2Ax+B$$

and it should be easy to determine the values of $A$ and $B$. But you are also given a nice point on the parabola, namely $(1,0)$, and this should tell you what $C$ is.

abiessu
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From the tangent line, we see that the parabola has y'(0)=3 at (0,6). This gives: $$y'(0)=2A(0)+B=3$$$$B=3$$ From y(0)=6, we get: $$y(0)=A(0)^2+B(0)+C=6$$ $$C=6$$ From y(1)=0, we get: $$y(1)=A(1)^2+3(1)+6=0$$ $$A+9=0$$ $$A=-9$$

reluctant mathematician
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