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Mathematics → Precalculus → Quadratic Equations → Finding Vertex of Parabola

Instructions for problem set  Find y-intercepts of a given quadratic equation

Questions


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

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

\(y = 3 x^2 -12 x + 11\)

\(y = -2 x^2 + 24 x -76\)

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

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

\(y = 3 x^2 -18 x + 24\)

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

\(y = 3 x^2 + 12 x + 15\)

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

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

\(y = -2 x^2 + 12 x -22\)

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

\(y = -2 x^2 -16 x -36\)

\(y = -2 x^2 -20 x -50\)

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

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

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

\(y = 3 x^2 + 18 x + 27\)

\(y = 3 x^2 + 6 x + 5\)

\(y = 2 x^2 -16 x + 28\)

\(y = 3 x^2 + 24 x + 45\)

\(y = -x^2 -12 x -35\)

\(y = -2 x^2 + 16 x -31\)

\(y = -2 x^2 -8 x -5\)

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

\(y = 3 x^2 + 12 x + 11\)

\(y = -2 x^2 -24 x -69\)

Answers


\(y = -x^2 + 2 x -4 \) ⇒ (\(1, -3)\)
\(y = -2 x^2 + 8 x -8 \) ⇒ (\(2, 0)\)
\(y = 3 x^2 -12 x + 11 \) ⇒ (\(2, -1)\)
\(y = -2 x^2 + 24 x -76 \) ⇒ (\(6, -4)\)

\(y = 3 x^2 -6 x + 6 \) ⇒ (\(1, 3)\)
\(y = x^2 -4 x + 4 \) ⇒ (\(2, 0)\)
\(y = 3 x^2 -18 x + 24 \) ⇒ (\(3, -3)\)
\(y = x^2 + 6 x + 8 \) ⇒ (\(-3, -1)\)

\(y = 3 x^2 + 12 x + 15 \) ⇒ (\(-2, 3)\)
\(y = -x^2 + 4 x -4 \) ⇒ (\(2, 0)\)
\(y = 3 x^2 + 6 x + 6 \) ⇒ (\(-1, 3)\)
\(y = -2 x^2 + 12 x -22 \) ⇒ (\(3, -4)\)

\(y = -2 x^2 -3 \) ⇒ (\(0, -3)\)
\(y = -2 x^2 -16 x -36 \) ⇒ (\(-4, -4)\)
\(y = -2 x^2 -20 x -50 \) ⇒ (\(-5, 0)\)
\(y = x^2 -4 x + 7 \) ⇒ (\(2, 3)\)

\(y = -x^2 -2 x -2 \) ⇒ (\(-1, -1)\)
\(y = x^2 -2 x + 3 \) ⇒ (\(1, 2)\)
\(y = 3 x^2 + 18 x + 27 \) ⇒ (\(-3, 0)\)
\(y = 3 x^2 + 6 x + 5 \) ⇒ (\(-1, 2)\)

\(y = 2 x^2 -16 x + 28 \) ⇒ (\(4, -4)\)
\(y = 3 x^2 + 24 x + 45 \) ⇒ (\(-4, -3)\)
\(y = -x^2 -12 x -35 \) ⇒ (\(-6, 1)\)
\(y = -2 x^2 + 16 x -31 \) ⇒ (\(4, 1)\)

\(y = -2 x^2 -8 x -5 \) ⇒ (\(-2, 3)\)
\(y = 2 x^2 + 8 x + 4 \) ⇒ (\(-2, -4)\)
\(y = 3 x^2 + 12 x + 11 \) ⇒ (\(-2, -1)\)
\(y = -2 x^2 -24 x -69 \) ⇒ (\(-6, 3)\)
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