At what point on the quadratic y= $x^{2}$ $x^2$ -4x-1 is the tangent parallel to the straight line y=-2x+5?

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Answer 1 · 2015-10-31T08:31:27

$(1, - 4)$ $(1, -4)$

$\frac{d}{d x} (x^{2} - 4 x - 1) = 2 x - 4$ $d/(dx) (x^2-4x-1) = 2x - 4$

$\frac{d}{d x} (- 2 x + 5) = - 2$ $d/(dx) (-2x+5) = -2$

So we are trying to find where these two slopes are equal, that is:

$2 x - 4 = - 2$ $2x - 4 = -2$

Hence $x = 1$ $x = 1$ , then $x^{2} - 4 x - 1 = - 4$ $x^2-4x-1 = -4$

graph{(y-x^2+3.92x+1)(y+2x-4.9)(y+2x+2)((x-1)^2+(y+4)^2-0.05) = 0 [-4.95, 9.29, -6.04, 1.076]}

At what point on the quadratic y=x2x^2-4x-1 is the tangent parallel to the straight line y=-2x+5?