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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


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

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

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

\(y = x^2 -10 x + 27\)

\(y = 3 x^2 + 36 x + 110\)

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

\(y = x^2 + 12 x + 37\)

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

\(y = -2 x^2 -20 x -54\)

\(y = -x^2 -10 x -25\)

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

\(y = 2 x^2 + 12 x + 16\)

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

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

\(y = -x^2 + 6 x -11\)

\(y = 3 x^2 -18 x + 30\)

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

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

\(y = 2 x^2 -24 x + 75\)

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

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

\(y = 2 x^2 + 8 x + 9\)

\(y = 3 x^2 -36 x + 104\)

\(y = 3 x^2 -18 x + 29\)

\(y = -2 x^2 -16 x -35\)

\(y = x^2 -12 x + 37\)

\(y = -2 x^2 + 12 x -16\)

\(y = -x^2 + 12 x -37\)

Answers


\(y = -2 x^2 + 12 x -17 \) ⇒ (\(3, 1)\)
\(y = x^2 -2 x + 3 \) ⇒ (\(1, 2)\)
\(y = -x^2 -4 x -8 \) ⇒ (\(-2, -4)\)
\(y = x^2 -10 x + 27 \) ⇒ (\(5, 2)\)

\(y = 3 x^2 + 36 x + 110 \) ⇒ (\(-6, 2)\)
\(y = 3 x^2 + 18 x + 24 \) ⇒ (\(-3, -3)\)
\(y = x^2 + 12 x + 37 \) ⇒ (\(-6, 1)\)
\(y = x^2 + 4 x + 4 \) ⇒ (\(-2, 0)\)

\(y = -2 x^2 -20 x -54 \) ⇒ (\(-5, -4)\)
\(y = -x^2 -10 x -25 \) ⇒ (\(-5, 0)\)
\(y = 3 x^2 + 12 x + 15 \) ⇒ (\(-2, 3)\)
\(y = 2 x^2 + 12 x + 16 \) ⇒ (\(-3, -2)\)

\(y = x^2 -6 x + 6 \) ⇒ (\(3, -3)\)
\(y = x^2 + 10 x + 22 \) ⇒ (\(-5, -3)\)
\(y = -x^2 + 6 x -11 \) ⇒ (\(3, -2)\)
\(y = 3 x^2 -18 x + 30 \) ⇒ (\(3, 3)\)

\(y = 2 x^2 -24 x + 68 \) ⇒ (\(6, -4)\)
\(y = x^2 -4 x + 4 \) ⇒ (\(2, 0)\)
\(y = 2 x^2 -24 x + 75 \) ⇒ (\(6, 3)\)
\(y = x^2 + 2 x + 4 \) ⇒ (\(-1, 3)\)

\(y = 3 x^2 + 24 x + 45 \) ⇒ (\(-4, -3)\)
\(y = 2 x^2 + 8 x + 9 \) ⇒ (\(-2, 1)\)
\(y = 3 x^2 -36 x + 104 \) ⇒ (\(6, -4)\)
\(y = 3 x^2 -18 x + 29 \) ⇒ (\(3, 2)\)

\(y = -2 x^2 -16 x -35 \) ⇒ (\(-4, -3)\)
\(y = x^2 -12 x + 37 \) ⇒ (\(6, 1)\)
\(y = -2 x^2 + 12 x -16 \) ⇒ (\(3, 2)\)
\(y = -x^2 + 12 x -37 \) ⇒ (\(6, -1)\)
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