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MathematicsPrecalculusQuadratic Equations → Finding Intercepts

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

Questions


\(y = x^2 -x -20\)

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

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

\(y = x^2 -36\)

\(y = x^2 + 7 x + 6\)

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

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

\(y = 4 x^2 + 9 x + 2\)

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

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

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

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

\(y = 4 x^2 -12 x\)

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

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

\(y = 2 x^2\)

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

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

\(y = 4 x^2 + 11 x -3\)

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

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

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

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

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

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

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

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

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

Answers


\(y = x^2 -x -20 \) ⇒
\(5, -4\)

\(y = x^2 + 6 x \) ⇒
\(-6, 0\)

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

\(y = x^2 -36 \) ⇒
\(-6, 6\)

\(y = x^2 + 7 x + 6 \) ⇒
\(-6, -1\)

\(y = x^2 -6 x + 5 \) ⇒
\(1, 5\)

\(y = x^2 + 10 x + 24 \) ⇒
\(-4, -6\)

\(y = 4 x^2 + 9 x + 2 \) ⇒
\(-1 / 4, -2\)

\(y = x^2 + x -6 \) ⇒
\(2, -3\)

\(y = 2 x^2 + 4 x + 2 \) ⇒
\(-1, -1\)

\(y = 2 x^2 -9 x + 9 \) ⇒
\(3 / 2, 3\)

\(y = 2 x^2 -3 x \) ⇒
\(3 / 2, 0\)

\(y = 4 x^2 -12 x \) ⇒
\(0, 3\)

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

\(y = x^2 + 2 x -3 \) ⇒
\(1, -3\)

\(y = 2 x^2 \) ⇒
\(0, 0\)

\(y = 4 x^2 -x -3 \) ⇒
\(-3 / 4, 1\)

\(y = x^2 -2 x -3 \) ⇒
\(3, -1\)

\(y = 4 x^2 + 11 x -3 \) ⇒
\(1 / 4, -3\)

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

\(y = 2 x^2 -4 x \) ⇒
\(0, 2\)

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

\(y = x^2 + x -2 \) ⇒
\(-2, 1\)

\(y = x^2 -2 x -24 \) ⇒
\(6, -4\)

\(y = x^2 + 2 x -15 \) ⇒
\(3, -5\)

\(y = x^2 -6 x + 9 \) ⇒
\(3, 3\)

\(y = x^2 -2 x -3 \) ⇒
\(-1, 3\)

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