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MathematicsPrecalculusQuadratic Equations → Finding Vertex of Parabola

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

Questions


\(y = 3 x^2 -30 x + 73\)

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

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

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

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

\(y = 3 x^2 + 36 x + 106\)

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

\(y = -2 x^2 + 16 x -36\)

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

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

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

\(y = x^2 + 1\)

\(y = -x^2\)

\(y = x^2 + 3\)

\(y = x^2 + 10 x + 21\)

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

\(y = 2 x^2 -8 x + 5\)

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

\(y = x^2 -12 x + 32\)

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

\(y = 3 x^2\)

\(y = 2 x^2\)

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

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

\(y = 3 x^2 + 36 x + 111\)

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

\(y = -x^2 + 8 x -13\)

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

Answers


\(y = 3 x^2 -30 x + 73 \) ⇒ (\(5, -2)\)

\(y = 2 x^2 + 1 \) ⇒ (\(0, 1)\)

\(y = -x^2 -6 x -12 \) ⇒ (\(-3, -3)\)

\(y = 2 x^2 -12 x + 20 \) ⇒ (\(3, 2)\)

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

\(y = 3 x^2 + 36 x + 106 \) ⇒ (\(-6, -2)\)

\(y = 2 x^2 -4 x -1 \) ⇒ (\(1, -3)\)

\(y = -2 x^2 + 16 x -36 \) ⇒ (\(4, -4)\)

\(y = x^2 -10 x + 24 \) ⇒ (\(5, -1)\)

\(y = -x^2 -2 x -4 \) ⇒ (\(-1, -3)\)

\(y = x^2 + 6 x + 11 \) ⇒ (\(-3, 2)\)

\(y = x^2 + 1 \) ⇒ (\(0, 1)\)

\(y = -x^2 \) ⇒ (\(0, 0)\)

\(y = x^2 + 3 \) ⇒ (\(0, 3)\)

\(y = x^2 + 10 x + 21 \) ⇒ (\(-5, -4)\)

\(y = 3 x^2 -1 \) ⇒ (\(0, -1)\)

\(y = 2 x^2 -8 x + 5 \) ⇒ (\(2, -3)\)

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

\(y = x^2 -12 x + 32 \) ⇒ (\(6, -4)\)

\(y = -x^2 + 10 x -22 \) ⇒ (\(5, 3)\)

\(y = 3 x^2 \) ⇒ (\(0, 0)\)

\(y = 2 x^2 \) ⇒ (\(0, 0)\)

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

\(y = -x^2 + 4 x -3 \) ⇒ (\(2, 1)\)

\(y = 3 x^2 + 36 x + 111 \) ⇒ (\(-6, 3)\)

\(y = 2 x^2 -4 x + 3 \) ⇒ (\(1, 1)\)

\(y = -x^2 + 8 x -13 \) ⇒ (\(4, 3)\)

\(y = x^2 + 2 x \) ⇒ (\(-1, -1)\)
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