Question

The graph of the quadratic function y=2x^2-4x+1 is pictured below along with the point P=(-1,7) on the parabola and the tangent line through P

Answer 1

The equation of the function passing through the given point is

y = -8x -1

The equation of a line in point-slope form is expressed as:

$y-y_0=m(x-x_0)$

m is the slope

$(x_0,y_0)$ is the point on the line

Given the equation $y=2x^2-4x+1$

Get the slope by differentiating at x = -1

$m=\frac{dy}{dx} =4x-4\\m=\frac{dy}{dx} =4(-1)-4\\m=\frac{dy}{dx} =-4-4\\m=\frac{dy}{dx} =-8$

Substitute (-1, 7) and m = -8 into the formula above to have $y-7=-8(x+1)$

Simplify

$y-7=-8x-8\\y+8x = -8+7\\y+8x=-1\\y=-8x-1$

Hence the equation of the function passing through the given point is

y = -8x -1

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