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MathematicsPrecalculusQuadratic Equations → Finding Intercepts

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

Questions


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

\(y = x^2 -9\)

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

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

\(y = x^2 -8 x + 12\)

\(y = 2 x^2 -5 x + 3\)

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

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

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

\(y = 3 x^2 -9 x + 6\)

\(y = x^2 + x\)

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

\(y = x^2\)

\(y = 3 x^2 -10 x + 3\)

\(y = 3 x^2 + 4 x + 1\)

\(y = 3 x^2 -7 x + 2\)

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

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

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

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

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

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

\(y = x^2 -x\)

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

\(y = x^2 + 10 x + 25\)

\(y = 4 x^2 + 7 x + 3\)

\(y = 3 x^2 -5 x -2\)

\(y = x^2 -36\)

Answers


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

\(y = x^2 -9 \) ⇒
\(-3, 3\)

\(y = x^2 + 9 x + 18 \) ⇒
\(-6, -3\)

\(y = 2 x^2 -6 x \) ⇒
\(0, 3\)

\(y = x^2 -8 x + 12 \) ⇒
\(6, 2\)

\(y = 2 x^2 -5 x + 3 \) ⇒
\(3 / 2, 1\)

\(y = x^2 -5 x + 6 \) ⇒
\(3, 2\)

\(y = x^2 + x -30 \) ⇒
\(5, -6\)

\(y = 3 x^2 -3 x \) ⇒
\(1, 0\)

\(y = 3 x^2 -9 x + 6 \) ⇒
\(1, 2\)

\(y = x^2 + x \) ⇒
\(-1, 0\)

\(y = 3 x^2 -6 x \) ⇒
\(0, 2\)

\(y = x^2 \) ⇒
\(0, 0\)

\(y = 3 x^2 -10 x + 3 \) ⇒
\(1 / 3, 3\)

\(y = 3 x^2 + 4 x + 1 \) ⇒
\(-1 / 3, -1\)

\(y = 3 x^2 -7 x + 2 \) ⇒
\(1 / 3, 2\)

\(y = 4 x^2 + 4 x \) ⇒
\(0, -1\)

\(y = x^2 -2 x \) ⇒
\(0, 2\)

\(y = x^2 -4 x \) ⇒
\(4, 0\)

\(y = 4 x^2 + 3 x \) ⇒
\(-3 / 4, 0\)

\(y = x^2 + 3 x -4 \) ⇒
\(1, -4\)

\(y = 3 x^2 + 6 x + 3 \) ⇒
\(-1, -1\)

\(y = x^2 -x \) ⇒
\(0, 1\)

\(y = 4 x^2 + x \) ⇒
\(-1 / 4, 0\)

\(y = x^2 + 10 x + 25 \) ⇒
\(-5, -5\)

\(y = 4 x^2 + 7 x + 3 \) ⇒
\(-3 / 4, -1\)

\(y = 3 x^2 -5 x -2 \) ⇒
\(-1 / 3, 2\)

\(y = x^2 -36 \) ⇒
\(6, -6\)
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