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MathematicsPrecalculusQuadratic Equations → Finding Vertex of Parabola

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

Questions


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

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

\(y = -x^2 -8 x -16\)

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

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

\(y = -2 x^2 + 20 x -52\)

\(y = 3 x^2 + 18 x + 26\)

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

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

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

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

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

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

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

\(y = x^2 -3\)

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

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

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

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

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

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

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

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

\(y = 2 x^2 + 4 x -1\)

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

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

\(y = -2 x^2 + 24 x -69\)

\(y = 3 x^2 + 12 x + 11\)

Answers


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

\(y = x^2 -2 x + 2 \) ⇒ (\(1, 1)\)

\(y = -x^2 -8 x -16 \) ⇒ (\(-4, 0)\)

\(y = -x^2 + 6 x -12 \) ⇒ (\(3, -3)\)

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

\(y = -2 x^2 + 20 x -52 \) ⇒ (\(5, -2)\)

\(y = 3 x^2 + 18 x + 26 \) ⇒ (\(-3, -1)\)

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

\(y = x^2 + 4 x + 5 \) ⇒ (\(-2, 1)\)

\(y = x^2 -12 x + 37 \) ⇒ (\(6, 1)\)

\(y = -x^2 -10 x -25 \) ⇒ (\(-5, 0)\)

\(y = -2 x^2 -8 x -7 \) ⇒ (\(-2, 1)\)

\(y = x^2 -2 x \) ⇒ (\(1, -1)\)

\(y = -x^2 + 4 x -6 \) ⇒ (\(2, -2)\)

\(y = x^2 -3 \) ⇒ (\(0, -3)\)

\(y = 3 x^2 + 3 \) ⇒ (\(0, 3)\)

\(y = x^2 -8 x + 18 \) ⇒ (\(4, 2)\)

\(y = x^2 + 2 x -2 \) ⇒ (\(-1, -3)\)

\(y = -x^2 -8 x -15 \) ⇒ (\(-4, 1)\)

\(y = -x^2 -2 x -2 \) ⇒ (\(-1, -1)\)

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

\(y = x^2 -10 x + 25 \) ⇒ (\(5, 0)\)

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

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

\(y = -2 x^2 -4 \) ⇒ (\(0, -4)\)

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

\(y = -2 x^2 + 24 x -69 \) ⇒ (\(6, 3)\)

\(y = 3 x^2 + 12 x + 11 \) ⇒ (\(-2, -1)\)
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