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MathematicsPrecalculusQuadratic Equations → Finding Intercepts

Instructions for problem set  Find x-intercepts of the given quadratic equation

Questions


\(y = 3 x^2 + x\)

\(y = x^2 + 9 x + 18\)

\(y = 3 x^2 -3 x -6\)

\(y = 3 x^2\)

\(y = x^2 -11 x + 30\)

\(y = x^2 + 8 x + 16\)

\(y = x^2 + 5 x + 6\)

\(y = x^2 -3 x + 2\)

\(y = 2 x^2 + 2 x\)

\(y = x^2 + 6 x + 8\)

\(y = 2 x^2 + 2 x -4\)

\(y = x^2 -3 x -18\)

\(y = 3 x^2 -6 x + 3\)

\(y = 4 x^2 -5 x + 1\)

\(y = 3 x^2 -8 x -3\)

\(y = x^2 -2 x -3\)

\(y = x^2 -10 x + 24\)

\(y = x^2 -5 x -6\)

\(y = 3 x^2 -6 x\)

\(y = x^2 -2 x -3\)

\(y = x^2 -2 x -8\)

\(y = 4 x^2 + 8 x\)

\(y = x^2 + x -6\)

\(y = x^2 -5 x + 6\)

\(y = 4 x^2 + 3 x\)

\(y = x^2 -2 x -15\)

\(y = x^2 -4 x + 3\)

\(y = x^2 -11 x + 30\)

Answers


\(y = 3 x^2 + x \) ⇒
\(-1 / 3, 0\)

\(y = x^2 + 9 x + 18 \) ⇒
\(-3, -6\)

\(y = 3 x^2 -3 x -6 \) ⇒
\(-1, 2\)

\(y = 3 x^2 \) ⇒
\(0, 0\)

\(y = x^2 -11 x + 30 \) ⇒
\(5, 6\)

\(y = x^2 + 8 x + 16 \) ⇒
\(-4, -4\)

\(y = x^2 + 5 x + 6 \) ⇒
\(-2, -3\)

\(y = x^2 -3 x + 2 \) ⇒
\(2, 1\)

\(y = 2 x^2 + 2 x \) ⇒
\(-1, 0\)

\(y = x^2 + 6 x + 8 \) ⇒
\(-4, -2\)

\(y = 2 x^2 + 2 x -4 \) ⇒
\(1, -2\)

\(y = x^2 -3 x -18 \) ⇒
\(-3, 6\)

\(y = 3 x^2 -6 x + 3 \) ⇒
\(1, 1\)

\(y = 4 x^2 -5 x + 1 \) ⇒
\(1 / 4, 1\)

\(y = 3 x^2 -8 x -3 \) ⇒
\(-1 / 3, 3\)

\(y = x^2 -2 x -3 \) ⇒
\(3, -1\)

\(y = x^2 -10 x + 24 \) ⇒
\(4, 6\)

\(y = x^2 -5 x -6 \) ⇒
\(-1, 6\)

\(y = 3 x^2 -6 x \) ⇒
\(0, 2\)

\(y = x^2 -2 x -3 \) ⇒
\(-1, 3\)

\(y = x^2 -2 x -8 \) ⇒
\(4, -2\)

\(y = 4 x^2 + 8 x \) ⇒
\(0, -2\)

\(y = x^2 + x -6 \) ⇒
\(2, -3\)

\(y = x^2 -5 x + 6 \) ⇒
\(2, 3\)

\(y = 4 x^2 + 3 x \) ⇒
\(-3 / 4, 0\)

\(y = x^2 -2 x -15 \) ⇒
\(-3, 5\)

\(y = x^2 -4 x + 3 \) ⇒
\(3, 1\)

\(y = x^2 -11 x + 30 \) ⇒
\(6, 5\)
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