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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


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

\(y = 2 x^2 -16 x + 33\)

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

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

\(y = x^2 + 10 x + 22\)

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

\(y = 2 x^2 + 20 x + 47\)

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

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

\(y = -2 x^2 -16 x -34\)

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

\(y = 3 x^2 -12 x + 8\)

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

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

\(y = 2 x^2 + 24 x + 68\)

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

\(y = 3 x^2 + 36 x + 109\)

\(y = 3 x^2 -18 x + 25\)

\(y = -2 x^2 -20 x -51\)

\(y = 3 x^2 -30 x + 72\)

\(y = 3 x^2 -36 x + 106\)

\(y = -2 x^2 + 24 x -71\)

\(y = 3 x^2 -24 x + 50\)

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

\(y = -x^2 -10 x -28\)

\(y = 2 x^2 -12 x + 17\)

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

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

Answers


\(y = 2 x^2 + 4 x -2 \) ⇒ (\(-1, -4)\)
\(y = 2 x^2 -16 x + 33 \) ⇒ (\(4, 1)\)
\(y = x^2 -6 x + 6 \) ⇒ (\(3, -3)\)
\(y = -2 x^2 -20 x -50 \) ⇒ (\(-5, 0)\)

\(y = x^2 + 10 x + 22 \) ⇒ (\(-5, -3)\)
\(y = -2 x^2 + 1 \) ⇒ (\(0, 1)\)
\(y = 2 x^2 + 20 x + 47 \) ⇒ (\(-5, -3)\)
\(y = -2 x^2 + 16 x -31 \) ⇒ (\(4, 1)\)

\(y = -2 x^2 -8 x -6 \) ⇒ (\(-2, 2)\)
\(y = -2 x^2 -16 x -34 \) ⇒ (\(-4, -2)\)
\(y = x^2 + 2 x + 3 \) ⇒ (\(-1, 2)\)
\(y = 3 x^2 -12 x + 8 \) ⇒ (\(2, -4)\)

\(y = x^2 -2 x -3 \) ⇒ (\(1, -4)\)
\(y = x^2 -4 x + 2 \) ⇒ (\(2, -2)\)
\(y = 2 x^2 + 24 x + 68 \) ⇒ (\(-6, -4)\)
\(y = -2 x^2 -4 x -2 \) ⇒ (\(-1, 0)\)

\(y = 3 x^2 + 36 x + 109 \) ⇒ (\(-6, 1)\)
\(y = 3 x^2 -18 x + 25 \) ⇒ (\(3, -2)\)
\(y = -2 x^2 -20 x -51 \) ⇒ (\(-5, -1)\)
\(y = 3 x^2 -30 x + 72 \) ⇒ (\(5, -3)\)

\(y = 3 x^2 -36 x + 106 \) ⇒ (\(6, -2)\)
\(y = -2 x^2 + 24 x -71 \) ⇒ (\(6, 1)\)
\(y = 3 x^2 -24 x + 50 \) ⇒ (\(4, 2)\)
\(y = 2 x^2 + 4 x \) ⇒ (\(-1, -2)\)

\(y = -x^2 -10 x -28 \) ⇒ (\(-5, -3)\)
\(y = 2 x^2 -12 x + 17 \) ⇒ (\(3, -1)\)
\(y = x^2 -2 x \) ⇒ (\(1, -1)\)
\(y = 3 x^2 + 18 x + 24 \) ⇒ (\(-3, -3)\)
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