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MathematicsPrecalculusQuadratic Equations → Finding Intercepts

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

Questions


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

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

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

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

\(y = x^2 + x\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(y = 3 x^2 -2 x -1\)

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

\(y = 4 x^2 -15 x + 9\)

\(y = x^2 + 7 x + 10\)

Answers


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

\(y = x^2 + 8 x + 12 \) ⇒
\(-6, -2\)

\(y = x^2 + 8 x + 12 \) ⇒
\(-2, -6\)

\(y = x^2 + 3 x \) ⇒
\(0, -3\)

\(y = x^2 + x \) ⇒
\(0, -1\)

\(y = x^2 -9 x + 18 \) ⇒
\(3, 6\)

\(y = x^2 -5 x + 4 \) ⇒
\(1, 4\)

\(y = x^2 -x -30 \) ⇒
\(-5, 6\)

\(y = x^2 -3 x -18 \) ⇒
\(6, -3\)

\(y = 4 x^2 + 12 x \) ⇒
\(0, -3\)

\(y = 4 x^2 + 9 x -9 \) ⇒
\(3 / 4, -3\)

\(y = x^2 -3 x -18 \) ⇒
\(-3, 6\)

\(y = x^2 + x -20 \) ⇒
\(-5, 4\)

\(y = x^2 -6 x \) ⇒
\(0, 6\)

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

\(y = 2 x^2 -4 x \) ⇒
\(0, 2\)

\(y = x^2 + 11 x + 30 \) ⇒
\(-6, -5\)

\(y = x^2 + 2 x -3 \) ⇒
\(1, -3\)

\(y = 4 x^2 + 4 x \) ⇒
\(0, -1\)

\(y = x^2 -6 x + 9 \) ⇒
\(3, 3\)

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

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

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

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

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

\(y = x^2 + 11 x + 30 \) ⇒
\(-5, -6\)

\(y = 4 x^2 -15 x + 9 \) ⇒
\(3 / 4, 3\)

\(y = x^2 + 7 x + 10 \) ⇒
\(-2, -5\)
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First Coefficient

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