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MathematicsPrecalculusQuadratic Equations → Finding Vertex of Parabola

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

Questions


\(y = 3 x^2 + 12 x + 10\)

\(y = x^2 + 6 x + 9\)

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

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

\(y = 3 x^2 -30 x + 76\)

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

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

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

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

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

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

\(y = x^2 -6 x + 7\)

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

\(y = 3 x^2 -6 x + 4\)

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

\(y = x^2 + 3\)

\(y = x^2 -4\)

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

\(y = -2 x^2 -8 x -10\)

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

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

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

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

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

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

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

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

\(y = x^2 -8 x + 15\)

Answers


\(y = 3 x^2 + 12 x + 10 \) ⇒ (\(-2, -2)\)

\(y = x^2 + 6 x + 9 \) ⇒ (\(-3, 0)\)

\(y = x^2 -12 x + 35 \) ⇒ (\(6, -1)\)

\(y = -2 x^2 -3 \) ⇒ (\(0, -3)\)

\(y = 3 x^2 -30 x + 76 \) ⇒ (\(5, 1)\)

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

\(y = -x^2 + 6 x -11 \) ⇒ (\(3, -2)\)

\(y = x^2 + 10 x + 24 \) ⇒ (\(-5, -1)\)

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

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

\(y = 2 x^2 + 4 x \) ⇒ (\(-1, -2)\)

\(y = x^2 -6 x + 7 \) ⇒ (\(3, -2)\)

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

\(y = 3 x^2 -6 x + 4 \) ⇒ (\(1, 1)\)

\(y = -x^2 + 4 x -5 \) ⇒ (\(2, -1)\)

\(y = x^2 + 3 \) ⇒ (\(0, 3)\)

\(y = x^2 -4 \) ⇒ (\(0, -4)\)

\(y = -x^2 + 2 x -5 \) ⇒ (\(1, -4)\)

\(y = -2 x^2 -8 x -10 \) ⇒ (\(-2, -2)\)

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

\(y = x^2 + 2 x + 2 \) ⇒ (\(-1, 1)\)

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

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

\(y = x^2 + 2 x -3 \) ⇒ (\(-1, -4)\)

\(y = -x^2 -10 x -22 \) ⇒ (\(-5, 3)\)

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

\(y = x^2 + 2 x -1 \) ⇒ (\(-1, -2)\)

\(y = x^2 -8 x + 15 \) ⇒ (\(4, -1)\)
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