Math Problems

MathematicsPrecalculusQuadratic Equations → Finding Vertex of Parabola

Find y-intercepts of a given quadratic equation

Questions

 $$y = 3 x^2 + 18 x + 26$$ $$y = 3 x^2 -12 x + 15$$ $$y = x^2 -12 x + 36$$ $$y = 3 x^2 + 6 x + 4$$ $$y = 3 x^2 + 12 x + 12$$ $$y = -2 x^2 + 4 x -3$$ $$y = -2 x^2 + 8 x -5$$ $$y = 3 x^2 -24 x + 49$$ $$y = 3 x^2 + 24 x + 44$$ $$y = -2 x^2 -4$$ $$y = x^2 + 3$$ $$y = -x^2 -10 x -23$$ $$y = -2 x^2 + 4 x -4$$ $$y = x^2 + 4 x + 2$$ $$y = x^2 -6 x + 5$$ $$y = -x^2 -8 x -13$$ $$y = 2 x^2 + 8 x + 8$$ $$y = -2 x^2 + 16 x -35$$ $$y = x^2 -8 x + 16$$ $$y = -x^2 + 8 x -19$$ $$y = -x^2 -2 x + 1$$ $$y = -x^2 -8 x -19$$ $$y = -2 x^2 + 3$$ $$y = 2 x^2 + 12 x + 17$$ $$y = 3 x^2 + 30 x + 75$$ $$y = x^2 + 2 x + 1$$ $$y = 2 x^2 -12 x + 21$$ $$y = x^2 + 8 x + 18$$

 $$y = 3 x^2 + 18 x + 26$$ ⇒ ($$-3, -1)$$ $$y = 3 x^2 -12 x + 15$$ ⇒ ($$2, 3)$$ $$y = x^2 -12 x + 36$$ ⇒ ($$6, 0)$$ $$y = 3 x^2 + 6 x + 4$$ ⇒ ($$-1, 1)$$ $$y = 3 x^2 + 12 x + 12$$ ⇒ ($$-2, 0)$$ $$y = -2 x^2 + 4 x -3$$ ⇒ ($$1, -1)$$ $$y = -2 x^2 + 8 x -5$$ ⇒ ($$2, 3)$$ $$y = 3 x^2 -24 x + 49$$ ⇒ ($$4, 1)$$ $$y = 3 x^2 + 24 x + 44$$ ⇒ ($$-4, -4)$$ $$y = -2 x^2 -4$$ ⇒ ($$0, -4)$$ $$y = x^2 + 3$$ ⇒ ($$0, 3)$$ $$y = -x^2 -10 x -23$$ ⇒ ($$-5, 2)$$ $$y = -2 x^2 + 4 x -4$$ ⇒ ($$1, -2)$$ $$y = x^2 + 4 x + 2$$ ⇒ ($$-2, -2)$$ $$y = x^2 -6 x + 5$$ ⇒ ($$3, -4)$$ $$y = -x^2 -8 x -13$$ ⇒ ($$-4, 3)$$ $$y = 2 x^2 + 8 x + 8$$ ⇒ ($$-2, 0)$$ $$y = -2 x^2 + 16 x -35$$ ⇒ ($$4, -3)$$ $$y = x^2 -8 x + 16$$ ⇒ ($$4, 0)$$ $$y = -x^2 + 8 x -19$$ ⇒ ($$4, -3)$$ $$y = -x^2 -2 x + 1$$ ⇒ ($$-1, 2)$$ $$y = -x^2 -8 x -19$$ ⇒ ($$-4, -3)$$ $$y = -2 x^2 + 3$$ ⇒ ($$0, 3)$$ $$y = 2 x^2 + 12 x + 17$$ ⇒ ($$-3, -1)$$ $$y = 3 x^2 + 30 x + 75$$ ⇒ ($$-5, 0)$$ $$y = x^2 + 2 x + 1$$ ⇒ ($$-1, 0)$$ $$y = 2 x^2 -12 x + 21$$ ⇒ ($$3, 3)$$ $$y = x^2 + 8 x + 18$$ ⇒ ($$-4, 2)$$