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Mathematics → Precalculus → Quadratic Equations → Finding Intercepts

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

Questions


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(y = 4 x^2 -10 x -6\)

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

\(y = x^2 + 7 x + 12\)

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

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

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

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

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

Answers


\(y = x^2 -x -30 \) ⇒
\(6, -5\)
\(y = x^2 -5 x + 4 \) ⇒
\(1, 4\)
\(y = x^2 -4 x + 3 \) ⇒
\(1, 3\)
\(y = x^2 -5 x + 4 \) ⇒
\(4, 1\)

\(y = 4 x^2 -3 x -1 \) ⇒
\(-1 / 4, 1\)
\(y = x^2 -8 x + 12 \) ⇒
\(6, 2\)
\(y = x^2 -7 x + 10 \) ⇒
\(2, 5\)
\(y = 4 x^2 + 2 x -2 \) ⇒
\(1 / 2, -1\)

\(y = x^2 + 3 x -10 \) ⇒
\(2, -5\)
\(y = x^2 + x -6 \) ⇒
\(2, -3\)
\(y = x^2 -x -12 \) ⇒
\(4, -3\)
\(y = 3 x^2 + 9 x \) ⇒
\(0, -3\)

\(y = x^2 + 6 x + 5 \) ⇒
\(-1, -5\)
\(y = 3 x^2 + 7 x -6 \) ⇒
\(2 / 3, -3\)
\(y = x^2 -4 x -5 \) ⇒
\(-1, 5\)
\(y = x^2 + 3 x -18 \) ⇒
\(-6, 3\)

\(y = x^2 -7 x + 6 \) ⇒
\(6, 1\)
\(y = 2 x^2 -9 x + 9 \) ⇒
\(3 / 2, 3\)
\(y = 4 x^2 -14 x + 6 \) ⇒
\(1 / 2, 3\)
\(y = x^2 -4 x -12 \) ⇒
\(-2, 6\)

\(y = 4 x^2 -10 x -6 \) ⇒
\(-1 / 2, 3\)
\(y = 3 x^2 + 3 x -6 \) ⇒
\(1, -2\)
\(y = x^2 + 7 x + 12 \) ⇒
\(-4, -3\)
\(y = x^2 + 4 x -5 \) ⇒
\(-5, 1\)

\(y = 2 x^2 -5 x -3 \) ⇒
\(-1 / 2, 3\)
\(y = 3 x^2 + 8 x -3 \) ⇒
\(1 / 3, -3\)
\(y = 4 x^2 -9 x -9 \) ⇒
\(-3 / 4, 3\)
\(y = x^2 + 8 x + 12 \) ⇒
\(-6, -2\)
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First Coefficient

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