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MathematicsPrecalculusQuadratic Equations → Finding Vertex of Parabola

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

Questions


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

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

\(y = x^2 + 8 x + 14\)

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

\(y = 2 x^2 + 24 x + 73\)

\(y = -2 x^2 -24 x -73\)

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

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

\(y = -x^2 -8 x -17\)

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

\(y = -x^2 -6 x -13\)

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

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

\(y = -2 x^2 + 16 x -30\)

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

\(y = 2 x^2 + 24 x + 72\)

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

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

\(y = 2 x^2 + 4 x + 5\)

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

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

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

\(y = -x^2 -10 x -29\)

\(y = x^2 -8 x + 17\)

\(y = 3 x^2 + 36 x + 107\)

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

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

\(y = -x^2 + 12 x -38\)

Answers


\(y = -2 x^2 + 16 x -32 \) ⇒ (\(4, 0)\)

\(y = -2 x^2 -8 x -9 \) ⇒ (\(-2, -1)\)

\(y = x^2 + 8 x + 14 \) ⇒ (\(-4, -2)\)

\(y = x^2 -4 x + 4 \) ⇒ (\(2, 0)\)

\(y = 2 x^2 + 24 x + 73 \) ⇒ (\(-6, 1)\)

\(y = -2 x^2 -24 x -73 \) ⇒ (\(-6, -1)\)

\(y = x^2 -6 x + 10 \) ⇒ (\(3, 1)\)

\(y = 2 x^2 -8 x + 8 \) ⇒ (\(2, 0)\)

\(y = -x^2 -8 x -17 \) ⇒ (\(-4, -1)\)

\(y = 2 x^2 -16 x + 31 \) ⇒ (\(4, -1)\)

\(y = -x^2 -6 x -13 \) ⇒ (\(-3, -4)\)

\(y = 2 x^2 -20 x + 50 \) ⇒ (\(5, 0)\)

\(y = -2 x^2 -20 x -51 \) ⇒ (\(-5, -1)\)

\(y = -2 x^2 + 16 x -30 \) ⇒ (\(4, 2)\)

\(y = 2 x^2 + 16 x + 32 \) ⇒ (\(-4, 0)\)

\(y = 2 x^2 + 24 x + 72 \) ⇒ (\(-6, 0)\)

\(y = -x^2 + 6 x -8 \) ⇒ (\(3, 1)\)

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

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

\(y = -2 x^2 -24 x -75 \) ⇒ (\(-6, -3)\)

\(y = x^2 + 4 x + 2 \) ⇒ (\(-2, -2)\)

\(y = -x^2 + 6 x -6 \) ⇒ (\(3, 3)\)

\(y = -x^2 -10 x -29 \) ⇒ (\(-5, -4)\)

\(y = x^2 -8 x + 17 \) ⇒ (\(4, 1)\)

\(y = 3 x^2 + 36 x + 107 \) ⇒ (\(-6, -1)\)

\(y = -x^2 -6 x -6 \) ⇒ (\(-3, 3)\)

\(y = -x^2 -8 x -18 \) ⇒ (\(-4, -2)\)

\(y = -x^2 + 12 x -38 \) ⇒ (\(6, -2)\)
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