calculus question (1 Viewer)

Relish9 · Aug 17, 2012

The curve y = ax² + bx + 8 has a minimum at (- 1, 4). Find 'a' and 'b'.

How do I solve a question like this?

dy/dx = 2ax + b

nightweaver066 · Aug 17, 2012

Since that point lies on the curve, sub it in to get one equation.

Since it has a minimum turning point at x = -1, the gradient must be 0 at that point, so sub x = -1 into dy/dx to get the second equation

Solve them two simultaneously to get a and b.

Carrotsticks · Aug 17, 2012

Relish9 said:
The curve y = ax² + bx + 8 has a minimum at (- 1, 4). Find 'a' and 'b'.

How do I solve a question like this?

dy/dx = 2ax + b

$\\ $Since the function passes through (-1,4), we can substitute that point in to acquire one equation:$ \\\\ 4 = a - b + 8 \\\\ \Rightarrow a-b = -4 \\\\ $Since (-1,4) is a stationary point, we can deduce that when $ x=-1, \frac{dy}{dx}=0 \\\\ -2a + b = 0 \Rightarrow 2a = b \\\\ $We can now solve them simultaneously:$ \\\\ a - 2a = -4 \\\\ \Rightarrow a =4 \\\\ b = 8$

shravan_872 · Aug 17, 2012

how did you the maths onto the forum, i wasnt able to do the same

calculus question (1 Viewer)

Relish9

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nightweaver066

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Carrotsticks

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shravan_872

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