Math Problems

MathematicsPrecalculusQuadratic Equations → Finding Intercepts

Find x-intercepts of the given quadratic equation

## Questions

 $$y = x^2 -4$$ $$y = x^2 + x -12$$ $$y = x^2 + 8 x + 12$$ $$y = 3 x^2 + 5 x + 2$$ $$y = x^2 -4 x$$ $$y = x^2 -7 x + 12$$ $$y = 3 x^2 + 9 x$$ $$y = 3 x^2 -9 x$$ $$y = 3 x^2 -3$$ $$y = x^2 -6 x$$ $$y = x^2 + 9 x + 20$$ $$y = 4 x^2 -10 x -6$$ $$y = x^2 + 2 x -8$$ $$y = x^2 -9 x + 18$$ $$y = x^2 + 9 x + 18$$ $$y = 2 x^2 + x -6$$ $$y = x^2 -3 x -18$$ $$y = 2 x^2 + 4 x$$ $$y = 2 x^2 -6 x$$ $$y = 4 x^2 + 5 x + 1$$ $$y = x^2 + 6 x + 5$$ $$y = 4 x^2 + 10 x -6$$ $$y = 4 x^2 -10 x + 4$$ $$y = 3 x^2 -5 x + 2$$ $$y = 2 x^2 -6 x + 4$$ $$y = x^2 + 8 x + 15$$ $$y = 4 x^2 -9 x + 2$$ $$y = 2 x^2 -3 x -2$$

## Answers

 $$y = x^2 -4$$ ⇒ $$2, -2$$ $$y = x^2 + x -12$$ ⇒ $$-4, 3$$ $$y = x^2 + 8 x + 12$$ ⇒ $$-2, -6$$ $$y = 3 x^2 + 5 x + 2$$ ⇒ $$-2 / 3, -1$$ $$y = x^2 -4 x$$ ⇒ $$0, 4$$ $$y = x^2 -7 x + 12$$ ⇒ $$4, 3$$ $$y = 3 x^2 + 9 x$$ ⇒ $$0, -3$$ $$y = 3 x^2 -9 x$$ ⇒ $$0, 3$$ $$y = 3 x^2 -3$$ ⇒ $$-1, 1$$ $$y = x^2 -6 x$$ ⇒ $$6, 0$$ $$y = x^2 + 9 x + 20$$ ⇒ $$-5, -4$$ $$y = 4 x^2 -10 x -6$$ ⇒ $$-1 / 2, 3$$ $$y = x^2 + 2 x -8$$ ⇒ $$-4, 2$$ $$y = x^2 -9 x + 18$$ ⇒ $$6, 3$$ $$y = x^2 + 9 x + 18$$ ⇒ $$-3, -6$$ $$y = 2 x^2 + x -6$$ ⇒ $$3 / 2, -2$$ $$y = x^2 -3 x -18$$ ⇒ $$6, -3$$ $$y = 2 x^2 + 4 x$$ ⇒ $$0, -2$$ $$y = 2 x^2 -6 x$$ ⇒ $$0, 3$$ $$y = 4 x^2 + 5 x + 1$$ ⇒ $$-1 / 4, -1$$ $$y = x^2 + 6 x + 5$$ ⇒ $$-1, -5$$ $$y = 4 x^2 + 10 x -6$$ ⇒ $$1 / 2, -3$$ $$y = 4 x^2 -10 x + 4$$ ⇒ $$1 / 2, 2$$ $$y = 3 x^2 -5 x + 2$$ ⇒ $$2 / 3, 1$$ $$y = 2 x^2 -6 x + 4$$ ⇒ $$1, 2$$ $$y = x^2 + 8 x + 15$$ ⇒ $$-3, -5$$ $$y = 4 x^2 -9 x + 2$$ ⇒ $$1 / 4, 2$$ $$y = 2 x^2 -3 x -2$$ ⇒ $$-1 / 2, 2$$
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