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MathematicsPrecalculusQuadratic Equations → Finding Vertex of Parabola

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

Questions


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

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

\(y = x^2 -12 x + 36\)

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

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

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

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

\(y = 3 x^2 -24 x + 49\)

\(y = 3 x^2 + 24 x + 44\)

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

\(y = x^2 + 3\)

\(y = -x^2 -10 x -23\)

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

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

\(y = x^2 -6 x + 5\)

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

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

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

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

\(y = -x^2 + 8 x -19\)

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

\(y = -x^2 -8 x -19\)

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

\(y = 2 x^2 + 12 x + 17\)

\(y = 3 x^2 + 30 x + 75\)

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

\(y = 2 x^2 -12 x + 21\)

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

Answers


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

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

\(y = x^2 -12 x + 36 \) ⇒ (\(6, 0)\)

\(y = 3 x^2 + 6 x + 4 \) ⇒ (\(-1, 1)\)

\(y = 3 x^2 + 12 x + 12 \) ⇒ (\(-2, 0)\)

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

\(y = -2 x^2 + 8 x -5 \) ⇒ (\(2, 3)\)

\(y = 3 x^2 -24 x + 49 \) ⇒ (\(4, 1)\)

\(y = 3 x^2 + 24 x + 44 \) ⇒ (\(-4, -4)\)

\(y = -2 x^2 -4 \) ⇒ (\(0, -4)\)

\(y = x^2 + 3 \) ⇒ (\(0, 3)\)

\(y = -x^2 -10 x -23 \) ⇒ (\(-5, 2)\)

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

\(y = x^2 + 4 x + 2 \) ⇒ (\(-2, -2)\)

\(y = x^2 -6 x + 5 \) ⇒ (\(3, -4)\)

\(y = -x^2 -8 x -13 \) ⇒ (\(-4, 3)\)

\(y = 2 x^2 + 8 x + 8 \) ⇒ (\(-2, 0)\)

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

\(y = x^2 -8 x + 16 \) ⇒ (\(4, 0)\)

\(y = -x^2 + 8 x -19 \) ⇒ (\(4, -3)\)

\(y = -x^2 -2 x + 1 \) ⇒ (\(-1, 2)\)

\(y = -x^2 -8 x -19 \) ⇒ (\(-4, -3)\)

\(y = -2 x^2 + 3 \) ⇒ (\(0, 3)\)

\(y = 2 x^2 + 12 x + 17 \) ⇒ (\(-3, -1)\)

\(y = 3 x^2 + 30 x + 75 \) ⇒ (\(-5, 0)\)

\(y = x^2 + 2 x + 1 \) ⇒ (\(-1, 0)\)

\(y = 2 x^2 -12 x + 21 \) ⇒ (\(3, 3)\)

\(y = x^2 + 8 x + 18 \) ⇒ (\(-4, 2)\)
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