Math Problems

MathematicsPrecalculusQuadratic Equations → Finding Vertex of Parabola

Find y-intercepts of a given quadratic equation

## Questions

 $$y = 3 x^2 -30 x + 76$$ $$y = x^2 + 12 x + 36$$ $$y = -2 x^2 -4 x -2$$ $$y = 3 x^2 + 18 x + 28$$ $$y = 2 x^2 -4 x + 5$$ $$y = -x^2 + 10 x -29$$ $$y = -x^2 + 4 x -6$$ $$y = 3 x^2 + 6 x + 6$$ $$y = -2 x^2 -4$$ $$y = 3 x^2 -30 x + 73$$ $$y = 2 x^2 + 24 x + 73$$ $$y = x^2 + 2 x + 1$$ $$y = -2 x^2 + 12 x -18$$ $$y = -x^2 + 8 x -14$$ $$y = -x^2 + 12 x -33$$ $$y = x^2 -1$$ $$y = -2 x^2 + 12 x -20$$ $$y = x^2 -8 x + 15$$ $$y = -2 x^2 + 16 x -34$$ $$y = 2 x^2 -20 x + 52$$ $$y = -2 x^2 + 16 x -29$$ $$y = -2 x^2 + 8 x -7$$ $$y = -x^2 -12 x -33$$ $$y = x^2 -2 x + 3$$ $$y = x^2 -10 x + 22$$ $$y = x^2 -8 x + 12$$ $$y = 3 x^2 + 24 x + 47$$ $$y = 3 x^2 -24 x + 49$$

 $$y = 3 x^2 -30 x + 76$$ ⇒ ($$5, 1)$$ $$y = x^2 + 12 x + 36$$ ⇒ ($$-6, 0)$$ $$y = -2 x^2 -4 x -2$$ ⇒ ($$-1, 0)$$ $$y = 3 x^2 + 18 x + 28$$ ⇒ ($$-3, 1)$$ $$y = 2 x^2 -4 x + 5$$ ⇒ ($$1, 3)$$ $$y = -x^2 + 10 x -29$$ ⇒ ($$5, -4)$$ $$y = -x^2 + 4 x -6$$ ⇒ ($$2, -2)$$ $$y = 3 x^2 + 6 x + 6$$ ⇒ ($$-1, 3)$$ $$y = -2 x^2 -4$$ ⇒ ($$0, -4)$$ $$y = 3 x^2 -30 x + 73$$ ⇒ ($$5, -2)$$ $$y = 2 x^2 + 24 x + 73$$ ⇒ ($$-6, 1)$$ $$y = x^2 + 2 x + 1$$ ⇒ ($$-1, 0)$$ $$y = -2 x^2 + 12 x -18$$ ⇒ ($$3, 0)$$ $$y = -x^2 + 8 x -14$$ ⇒ ($$4, 2)$$ $$y = -x^2 + 12 x -33$$ ⇒ ($$6, 3)$$ $$y = x^2 -1$$ ⇒ ($$0, -1)$$ $$y = -2 x^2 + 12 x -20$$ ⇒ ($$3, -2)$$ $$y = x^2 -8 x + 15$$ ⇒ ($$4, -1)$$ $$y = -2 x^2 + 16 x -34$$ ⇒ ($$4, -2)$$ $$y = 2 x^2 -20 x + 52$$ ⇒ ($$5, 2)$$ $$y = -2 x^2 + 16 x -29$$ ⇒ ($$4, 3)$$ $$y = -2 x^2 + 8 x -7$$ ⇒ ($$2, 1)$$ $$y = -x^2 -12 x -33$$ ⇒ ($$-6, 3)$$ $$y = x^2 -2 x + 3$$ ⇒ ($$1, 2)$$ $$y = x^2 -10 x + 22$$ ⇒ ($$5, -3)$$ $$y = x^2 -8 x + 12$$ ⇒ ($$4, -4)$$ $$y = 3 x^2 + 24 x + 47$$ ⇒ ($$-4, -1)$$ $$y = 3 x^2 -24 x + 49$$ ⇒ ($$4, 1)$$
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