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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

Sagot :

abidemiokin

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:

[tex]y-y_0=m(x-x_0)[/tex]

m is the slope

[tex](x_0,y_0)[/tex] is the point on the line

Given the equation [tex]y=2x^2-4x+1[/tex]

Get the slope by differentiating at x = -1

[tex]m=\frac{dy}{dx} =4x-4\\m=\frac{dy}{dx} =4(-1)-4\\m=\frac{dy}{dx} =-4-4\\m=\frac{dy}{dx} =-8[/tex]

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

Simplify

[tex]y-7=-8x-8\\y+8x = -8+7\\y+8x=-1\\y=-8x-1[/tex]

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

y = -8x -1

Learn more here: https://brainly.com/question/13723076

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