When b varies | Conic Section

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Saturday 6 October 2012

When b varies

The x-coordinate at the vertex is $\text{[math]}$ , which is found by differentiating the original equation $\text{[math]}$ , setting the resulting $dy/dx=2ax+b$ equal to zero (a critical point), and solving for $x$ . Substitute this x-coordinate into the original equation to yield:

$y=a\left (-\frac{b}{2a}\right )^2 + b \left ( -\frac{b}{2a} \right ) + c.$

Simplifying:

$=\frac{ab^2}{4a^2} -\frac{b^2}{2a} + c$

$=\frac{b^2}{4a} -\frac{2\cdot b^2}{2\cdot 2a} + c\cdot\frac{4a}{4a}$

$=\frac{-b^2+4ac}{4a}$

$=-\frac{b^2-4ac}{4a}=-\frac{D}{4a}.$

Thus, the vertex is at point

$\left (-\frac{b}{2a},-\frac{D}{4a}\right ).$

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