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Mathematics → Precalculus → Quadratic Equations → Finding Intercepts

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

Questions


\(y = 4 x^2 + 8 x\)

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

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

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

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

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

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

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

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

\(y = x^2 + 6 x + 8\)

\(y = x^2 -8 x + 16\)

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

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

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

\(y = 4 x^2 -8 x\)

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

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

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

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

\(y = x^2 -1\)

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

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

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

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

\(y = x^2 + 12 x + 36\)

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

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

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

Answers


\(y = 4 x^2 + 8 x \) ⇒
\(0, -2\)
\(y = x^2 + 9 x + 18 \) ⇒
\(-6, -3\)
\(y = 2 x^2 -6 x \) ⇒
\(0, 3\)
\(y = 4 x^2 -5 x + 1 \) ⇒
\(1 / 4, 1\)

\(y = 3 x^2 -6 x + 3 \) ⇒
\(1, 1\)
\(y = x^2 -x -20 \) ⇒
\(5, -4\)
\(y = x^2 -3 x -18 \) ⇒
\(6, -3\)
\(y = x^2 -7 x + 10 \) ⇒
\(5, 2\)

\(y = x^2 + 5 x + 6 \) ⇒
\(-3, -2\)
\(y = x^2 + 6 x + 8 \) ⇒
\(-2, -4\)
\(y = x^2 -8 x + 16 \) ⇒
\(4, 4\)
\(y = x^2 -3 x \) ⇒
\(0, 3\)

\(y = x^2 + 2 x -3 \) ⇒
\(-3, 1\)
\(y = 3 x^2 + 5 x -2 \) ⇒
\(1 / 3, -2\)
\(y = 4 x^2 -8 x \) ⇒
\(0, 2\)
\(y = 3 x^2 -x -2 \) ⇒
\(-2 / 3, 1\)

\(y = 2 x^2 -7 x + 3 \) ⇒
\(1 / 2, 3\)
\(y = 4 x^2 -3 x \) ⇒
\(3 / 4, 0\)
\(y = 4 x^2 -14 x + 6 \) ⇒
\(1 / 2, 3\)
\(y = x^2 -1 \) ⇒
\(1, -1\)

\(y = x^2 -2 x -15 \) ⇒
\(5, -3\)
\(y = 3 x^2 -2 x -1 \) ⇒
\(-1 / 3, 1\)
\(y = x^2 + 2 x -8 \) ⇒
\(-4, 2\)
\(y = x^2 + 8 x + 12 \) ⇒
\(-6, -2\)

\(y = x^2 + 12 x + 36 \) ⇒
\(-6, -6\)
\(y = 4 x^2 + 5 x + 1 \) ⇒
\(-1 / 4, -1\)
\(y = 2 x^2 + 2 x \) ⇒
\(0, -1\)
\(y = 3 x^2 -7 x + 2 \) ⇒
\(1 / 3, 2\)
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