【人気ダウンロード！】 y=x^2-4x 1 parabola 114384-La retta tangente alla parabola y=x^2+4x-1

As $\displaystyle y = \frac {x^2 2x 4}{2x^2 4x 9} \implies 2y = \frac {2x^2 4x 9 1}{2x^2 4x 9}$ Thus, $\displaystyle 2y = 1\frac {1}{2(x 1)^2 7} $ Squares can never be less than zero so the minmum value of the function $\displaystyle 2(x 1)^2 7 $ would be $7$ , or Maximum value of $\displaystyle \frac {1}{2(x 1

La retta tangente alla parabola y=x^2+4x-1- Finding the yintercept of a parabola can be tricky Although the yintercept is hidden, it does exist Use the equation of the function to find the yintercept y = 12x 2 48x 49 The yintercept has two parts the xvalue and the yvalue Note that the xvalue is always zero So, plug in zero for x and solve for y y = 12(0) 2 48(0) 49Let P be a variable point on the parabola y = 4x 2 1 Then, the locus of the midpoint of the point P and the foot of the perpendicular drawn from the point P to the line y = x is Then, the locus of the midpoint of the point P and the foot of the perpendicular