Math Problems

MathematicsPrecalculusQuadratic Equations → Finding Vertex of Parabola

Find y-intercepts of a given quadratic equation

## Questions

 $$y = 2 x^2 + 4 x -2$$ $$y = 2 x^2 -16 x + 33$$ $$y = x^2 -6 x + 6$$ $$y = -2 x^2 -20 x -50$$ $$y = x^2 + 10 x + 22$$ $$y = -2 x^2 + 1$$ $$y = 2 x^2 + 20 x + 47$$ $$y = -2 x^2 + 16 x -31$$ $$y = -2 x^2 -8 x -6$$ $$y = -2 x^2 -16 x -34$$ $$y = x^2 + 2 x + 3$$ $$y = 3 x^2 -12 x + 8$$ $$y = x^2 -2 x -3$$ $$y = x^2 -4 x + 2$$ $$y = 2 x^2 + 24 x + 68$$ $$y = -2 x^2 -4 x -2$$ $$y = 3 x^2 + 36 x + 109$$ $$y = 3 x^2 -18 x + 25$$ $$y = -2 x^2 -20 x -51$$ $$y = 3 x^2 -30 x + 72$$ $$y = 3 x^2 -36 x + 106$$ $$y = -2 x^2 + 24 x -71$$ $$y = 3 x^2 -24 x + 50$$ $$y = 2 x^2 + 4 x$$ $$y = -x^2 -10 x -28$$ $$y = 2 x^2 -12 x + 17$$ $$y = x^2 -2 x$$ $$y = 3 x^2 + 18 x + 24$$

 $$y = 2 x^2 + 4 x -2$$ ⇒ ($$-1, -4)$$ $$y = 2 x^2 -16 x + 33$$ ⇒ ($$4, 1)$$ $$y = x^2 -6 x + 6$$ ⇒ ($$3, -3)$$ $$y = -2 x^2 -20 x -50$$ ⇒ ($$-5, 0)$$ $$y = x^2 + 10 x + 22$$ ⇒ ($$-5, -3)$$ $$y = -2 x^2 + 1$$ ⇒ ($$0, 1)$$ $$y = 2 x^2 + 20 x + 47$$ ⇒ ($$-5, -3)$$ $$y = -2 x^2 + 16 x -31$$ ⇒ ($$4, 1)$$ $$y = -2 x^2 -8 x -6$$ ⇒ ($$-2, 2)$$ $$y = -2 x^2 -16 x -34$$ ⇒ ($$-4, -2)$$ $$y = x^2 + 2 x + 3$$ ⇒ ($$-1, 2)$$ $$y = 3 x^2 -12 x + 8$$ ⇒ ($$2, -4)$$ $$y = x^2 -2 x -3$$ ⇒ ($$1, -4)$$ $$y = x^2 -4 x + 2$$ ⇒ ($$2, -2)$$ $$y = 2 x^2 + 24 x + 68$$ ⇒ ($$-6, -4)$$ $$y = -2 x^2 -4 x -2$$ ⇒ ($$-1, 0)$$ $$y = 3 x^2 + 36 x + 109$$ ⇒ ($$-6, 1)$$ $$y = 3 x^2 -18 x + 25$$ ⇒ ($$3, -2)$$ $$y = -2 x^2 -20 x -51$$ ⇒ ($$-5, -1)$$ $$y = 3 x^2 -30 x + 72$$ ⇒ ($$5, -3)$$ $$y = 3 x^2 -36 x + 106$$ ⇒ ($$6, -2)$$ $$y = -2 x^2 + 24 x -71$$ ⇒ ($$6, 1)$$ $$y = 3 x^2 -24 x + 50$$ ⇒ ($$4, 2)$$ $$y = 2 x^2 + 4 x$$ ⇒ ($$-1, -2)$$ $$y = -x^2 -10 x -28$$ ⇒ ($$-5, -3)$$ $$y = 2 x^2 -12 x + 17$$ ⇒ ($$3, -1)$$ $$y = x^2 -2 x$$ ⇒ ($$1, -1)$$ $$y = 3 x^2 + 18 x + 24$$ ⇒ ($$-3, -3)$$