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Mathematics → Precalculus → Quadratic Equations → Finding Intercepts

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

Questions


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

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

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

\(y = 2 x^2\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Answers


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

\(y = x^2 -11 x + 30 \) ⇒
\(5, 6\)
\(y = 2 x^2 + 3 x \) ⇒
\(-3 / 2, 0\)
\(y = x^2 + x -20 \) ⇒
\(4, -5\)
\(y = 2 x^2 + 9 x + 9 \) ⇒
\(-3 / 2, -3\)

\(y = x^2 + 8 x + 12 \) ⇒
\(-6, -2\)
\(y = 2 x^2 + 7 x + 3 \) ⇒
\(-1 / 2, -3\)
\(y = 3 x^2 -3 x \) ⇒
\(1, 0\)
\(y = x^2 + 3 x -4 \) ⇒
\(-4, 1\)

\(y = x^2 + 7 x + 10 \) ⇒
\(-5, -2\)
\(y = x^2 + 9 x + 18 \) ⇒
\(-3, -6\)
\(y = 2 x^2 + 5 x + 2 \) ⇒
\(-1 / 2, -2\)
\(y = x^2 + 5 x + 4 \) ⇒
\(-1, -4\)

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

\(y = x^2 + 2 x + 1 \) ⇒
\(-1, -1\)
\(y = 4 x^2 + 9 x -9 \) ⇒
\(3 / 4, -3\)
\(y = x^2 -8 x + 15 \) ⇒
\(5, 3\)
\(y = x^2 + 8 x + 12 \) ⇒
\(-2, -6\)

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