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MathematicsPrecalculusQuadratic Equations → Finding Intercepts

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

Questions


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

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

\(y = 2 x^2 -x -1\)

\(y = 4 x^2 -14 x + 6\)

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

\(y = 2 x^2 -6 x + 4\)

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

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

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

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

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

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

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

\(y = 4 x^2 + 6 x -4\)

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

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

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

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

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

\(y = 4 x^2 + 10 x + 4\)

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

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

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

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

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

\(y = x^2 -36\)

\(y = x^2 -9 x + 20\)

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

Answers


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

\(y = x^2 + 4 x + 3 \) ⇒
\(-3, -1\)

\(y = 2 x^2 -x -1 \) ⇒
\(-1 / 2, 1\)

\(y = 4 x^2 -14 x + 6 \) ⇒
\(1 / 2, 3\)

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

\(y = 2 x^2 -6 x + 4 \) ⇒
\(1, 2\)

\(y = x^2 -6 x + 5 \) ⇒
\(5, 1\)

\(y = x^2 + 2 x -8 \) ⇒
\(-4, 2\)

\(y = 4 x^2 + 12 x \) ⇒
\(0, -3\)

\(y = 4 x^2 -3 x \) ⇒
\(3 / 4, 0\)

\(y = x^2 + 5 x + 4 \) ⇒
\(-1, -4\)

\(y = 4 x^2 -2 x \) ⇒
\(1 / 2, 0\)

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

\(y = 4 x^2 + 6 x -4 \) ⇒
\(1 / 2, -2\)

\(y = x^2 -x -30 \) ⇒
\(6, -5\)

\(y = x^2 + 7 x + 6 \) ⇒
\(-1, -6\)

\(y = x^2 + 10 x + 24 \) ⇒
\(-4, -6\)

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

\(y = 2 x^2 -7 x + 6 \) ⇒
\(3 / 2, 2\)

\(y = 4 x^2 + 10 x + 4 \) ⇒
\(-1 / 2, -2\)

\(y = x^2 + x -30 \) ⇒
\(-6, 5\)

\(y = x^2 + 3 x -10 \) ⇒
\(-5, 2\)

\(y = x^2 + 9 x + 18 \) ⇒
\(-3, -6\)

\(y = x^2 -3 x -18 \) ⇒
\(-3, 6\)

\(y = x^2 + 2 x -15 \) ⇒
\(3, -5\)

\(y = x^2 -36 \) ⇒
\(-6, 6\)

\(y = x^2 -9 x + 20 \) ⇒
\(4, 5\)

\(y = x^2 -x -6 \) ⇒
\(-2, 3\)
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First Coefficient

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