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3 years ago

I need to graph a parabola. I need to find the vertex and a couple coordinates. y=-x^2+2x-4

Mathematics

1 answer:

PolarNik [594]3 years ago

6 0

Hello :
y=-x²+2x-4
y = -(x²-2x)-4
y = - (x²-2x+1)+1-4
y = - (x-1)² -3
the vertex is : (1 , -3 )

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Step-by-step explanation:

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Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2

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(a) $f'(x)=-\frac{2}{x^3}$

(b) $y=-0.25x+0.75$

Step-by-step explanation:

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The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

$m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}$

Substitute x=2 in equation 2.

$f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25$

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

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$y=-0.25x+0.75$

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