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Mathematics → Precalculus → Quadratic Equations → Finding Intercepts

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

Questions


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(y = x^2 -1\)

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

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

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

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

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

Answers


\(y = 4 x^2 -x -3 \) ⇒
\(-3 / 4, 1\)
\(y = x^2 + x -2 \) ⇒
\(1, -2\)
\(y = 4 x^2 -x \) ⇒
\(1 / 4, 0\)
\(y = 3 x^2 + 9 x + 6 \) ⇒
\(-1, -2\)

\(y = x^2 -3 x -10 \) ⇒
\(5, -2\)
\(y = x^2 -5 x \) ⇒
\(0, 5\)
\(y = x^2 + x -30 \) ⇒
\(5, -6\)
\(y = 3 x^2 -3 x -6 \) ⇒
\(-1, 2\)

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

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

\(y = x^2 -x -2 \) ⇒
\(2, -1\)
\(y = x^2 -x -20 \) ⇒
\(-4, 5\)
\(y = x^2 -x -6 \) ⇒
\(3, -2\)
\(y = x^2 + 7 x + 12 \) ⇒
\(-3, -4\)

\(y = 2 x^2 + 3 x + 1 \) ⇒
\(-1 / 2, -1\)
\(y = x^2 -2 x -8 \) ⇒
\(4, -2\)
\(y = x^2 -1 \) ⇒
\(1, -1\)
\(y = x^2 + 9 x + 20 \) ⇒
\(-4, -5\)

\(y = x^2 + 3 x -18 \) ⇒
\(3, -6\)
\(y = x^2 -3 x \) ⇒
\(3, 0\)
\(y = 2 x^2 -5 x -3 \) ⇒
\(-1 / 2, 3\)
\(y = x^2 + 7 x + 10 \) ⇒
\(-5, -2\)
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