[最も選択された] parabola congruent to y=x^2 130969-Parabola congruent to y=x^2

Graph the parabola, y =x^21 by finding the turning point and using a table to find values for x and yNew parabola should be congruent (the same shape and size) to y = x2, with the same vertex, except it should open downward so its vertex will be its highest point Record the Find a way to change the equation to make the y = x2 parabola move 3 units to the left and stretch vertically, as in part (c)Your new parabola might look like y = 4x210 CHAPTER 16 PARTIAL DERIVATIVES Figure 1617 often more useful to sketch a few of its level curves than to sketch that surface Each level curve is the projection of a

Vertex Form

Parabola congruent to y=x^2

[最も共有された！ √] y=x^2 2x-8 in factored form 255494

 The factors of 5x2 5x −3 are x − x1 and x − x2, so we have the factored form y = (x − −5 √85 10)(x − −5 −√85 10) If you wanted vertex form, here is the answer y = 5(x2 xxxxxxx) − 3 y = 5(x2 x 1 4) − 3 − 5 4 y = 5(x − 1 2)2 −