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MathematicsPrecalculusQuadratic Equations → Finding Intercepts

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

Questions


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

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

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

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

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

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

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

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

\(y = 3 x^2\)

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

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

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

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

\(y = 3 x^2 + 12 x + 9\)

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

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

\(y = x^2 -1\)

\(y = x^2 -16\)

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

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

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

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

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

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

\(y = x^2 + x\)

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

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

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

Answers


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

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

\(y = 3 x^2 -3 x \) ⇒
\(0, 1\)

\(y = x^2 + 3 x -10 \) ⇒
\(2, -5\)

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

\(y = x^2 + x -20 \) ⇒
\(-5, 4\)

\(y = x^2 -8 x + 12 \) ⇒
\(6, 2\)

\(y = x^2 -6 x \) ⇒
\(6, 0\)

\(y = 3 x^2 \) ⇒
\(0, 0\)

\(y = 3 x^2 + 6 x \) ⇒
\(0, -2\)

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

\(y = x^2 + 2 x -15 \) ⇒
\(3, -5\)

\(y = x^2 + 2 x + 1 \) ⇒
\(-1, -1\)

\(y = 3 x^2 + 12 x + 9 \) ⇒
\(-1, -3\)

\(y = 3 x^2 + 9 x \) ⇒
\(0, -3\)

\(y = x^2 + 6 x + 9 \) ⇒
\(-3, -3\)

\(y = x^2 -1 \) ⇒
\(1, -1\)

\(y = x^2 -16 \) ⇒
\(4, -4\)

\(y = x^2 -6 x + 9 \) ⇒
\(3, 3\)

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

\(y = 4 x^2 -7 x -2 \) ⇒
\(-1 / 4, 2\)

\(y = x^2 -3 x -10 \) ⇒
\(-2, 5\)

\(y = 2 x^2 -3 x + 1 \) ⇒
\(1 / 2, 1\)

\(y = 4 x^2 + x -3 \) ⇒
\(3 / 4, -1\)

\(y = x^2 + x \) ⇒
\(-1, 0\)

\(y = x^2 -3 x \) ⇒
\(0, 3\)

\(y = 3 x^2 + 3 x \) ⇒
\(0, -1\)

\(y = x^2 + 5 x \) ⇒
\(0, -5\)
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