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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


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

\(y = -x^2 + 12 x -34\)

\(y = -x^2 + 6 x -10\)

\(y = 2 x^2 + 16 x + 32\)

\(y = -2 x^2 + 12 x -20\)

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

\(y = -x^2 + 10 x -24\)

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

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

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

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

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

\(y = -x^2 -8 x -18\)

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

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

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

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

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

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

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

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

\(y = x^2 -1\)

\(y = 3 x^2 + 24 x + 51\)

\(y = 3 x^2 + 6 x -1\)

\(y = 2 x^2 -8 x + 7\)

\(y = -2 x^2 + 24 x -70\)

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

\(y = 3 x^2 -6 x -1\)

Answers


\(y = -2 x^2 + 12 x -15 \) ⇒ (\(3, 3)\)
\(y = -x^2 + 12 x -34 \) ⇒ (\(6, 2)\)
\(y = -x^2 + 6 x -10 \) ⇒ (\(3, -1)\)
\(y = 2 x^2 + 16 x + 32 \) ⇒ (\(-4, 0)\)

\(y = -2 x^2 + 12 x -20 \) ⇒ (\(3, -2)\)
\(y = 3 x^2 + 18 x + 24 \) ⇒ (\(-3, -3)\)
\(y = -x^2 + 10 x -24 \) ⇒ (\(5, 1)\)
\(y = 3 x^2 + 36 x + 109 \) ⇒ (\(-6, 1)\)

\(y = -x^2 -2 x -5 \) ⇒ (\(-1, -4)\)
\(y = -x^2 -2 x -2 \) ⇒ (\(-1, -1)\)
\(y = 2 x^2 + 16 x + 28 \) ⇒ (\(-4, -4)\)
\(y = 3 x^2 -30 x + 72 \) ⇒ (\(5, -3)\)

\(y = -x^2 -8 x -18 \) ⇒ (\(-4, -2)\)
\(y = x^2 -4 x + 7 \) ⇒ (\(2, 3)\)
\(y = 2 x^2 -16 x + 35 \) ⇒ (\(4, 3)\)
\(y = -x^2 -6 x -11 \) ⇒ (\(-3, -2)\)

\(y = -x^2 + 4 x -4 \) ⇒ (\(2, 0)\)
\(y = -2 x^2 + 8 x -9 \) ⇒ (\(2, -1)\)
\(y = 3 x^2 + 12 x + 12 \) ⇒ (\(-2, 0)\)
\(y = x^2 + 12 x + 37 \) ⇒ (\(-6, 1)\)

\(y = -x^2 + 10 x -27 \) ⇒ (\(5, -2)\)
\(y = x^2 -1 \) ⇒ (\(0, -1)\)
\(y = 3 x^2 + 24 x + 51 \) ⇒ (\(-4, 3)\)
\(y = 3 x^2 + 6 x -1 \) ⇒ (\(-1, -4)\)

\(y = 2 x^2 -8 x + 7 \) ⇒ (\(2, -1)\)
\(y = -2 x^2 + 24 x -70 \) ⇒ (\(6, 2)\)
\(y = x^2 -12 x + 35 \) ⇒ (\(6, -1)\)
\(y = 3 x^2 -6 x -1 \) ⇒ (\(1, -4)\)
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