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Mathematics → Precalculus → Quadratic Equations → Finding Intercepts

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

Questions


\(y = x^2 -4\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(y = x^2 + 8 x + 15\)

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

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

Answers


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

\(y = x^2 -4 x \) ⇒
\(0, 4\)
\(y = x^2 -7 x + 12 \) ⇒
\(4, 3\)
\(y = 3 x^2 + 9 x \) ⇒
\(0, -3\)
\(y = 3 x^2 -9 x \) ⇒
\(0, 3\)

\(y = 3 x^2 -3 \) ⇒
\(-1, 1\)
\(y = x^2 -6 x \) ⇒
\(6, 0\)
\(y = x^2 + 9 x + 20 \) ⇒
\(-5, -4\)
\(y = 4 x^2 -10 x -6 \) ⇒
\(-1 / 2, 3\)

\(y = x^2 + 2 x -8 \) ⇒
\(-4, 2\)
\(y = x^2 -9 x + 18 \) ⇒
\(6, 3\)
\(y = x^2 + 9 x + 18 \) ⇒
\(-3, -6\)
\(y = 2 x^2 + x -6 \) ⇒
\(3 / 2, -2\)

\(y = x^2 -3 x -18 \) ⇒
\(6, -3\)
\(y = 2 x^2 + 4 x \) ⇒
\(0, -2\)
\(y = 2 x^2 -6 x \) ⇒
\(0, 3\)
\(y = 4 x^2 + 5 x + 1 \) ⇒
\(-1 / 4, -1\)

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

\(y = 2 x^2 -6 x + 4 \) ⇒
\(1, 2\)
\(y = x^2 + 8 x + 15 \) ⇒
\(-3, -5\)
\(y = 4 x^2 -9 x + 2 \) ⇒
\(1 / 4, 2\)
\(y = 2 x^2 -3 x -2 \) ⇒
\(-1 / 2, 2\)
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