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MathematicsPrecalculusQuadratic Equations → Finding Vertex of Parabola

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

Questions


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

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

\(y = x^2 -4\)

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

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

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

\(y = -2 x^2 -20 x -53\)

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

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

\(y = -2 x^2 + 12 x -18\)

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

\(y = 3 x^2 + 18 x + 29\)

\(y = -x^2 -4 x -8\)

\(y = -x^2 -8 x -20\)

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

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

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

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

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

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

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

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

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

\(y = -x^2 + 8 x -20\)

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

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

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

\(y = 3 x^2 + 30 x + 71\)

Answers


\(y = -x^2 + 8 x -13 \) ⇒ (\(4, 3)\)

\(y = x^2 -10 x + 22 \) ⇒ (\(5, -3)\)

\(y = x^2 -4 \) ⇒ (\(0, -4)\)

\(y = -x^2 -2 x + 1 \) ⇒ (\(-1, 2)\)

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

\(y = 2 x^2 -2 \) ⇒ (\(0, -2)\)

\(y = -2 x^2 -20 x -53 \) ⇒ (\(-5, -3)\)

\(y = x^2 -6 x + 12 \) ⇒ (\(3, 3)\)

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

\(y = -2 x^2 + 12 x -18 \) ⇒ (\(3, 0)\)

\(y = x^2 -10 x + 23 \) ⇒ (\(5, -2)\)

\(y = 3 x^2 + 18 x + 29 \) ⇒ (\(-3, 2)\)

\(y = -x^2 -4 x -8 \) ⇒ (\(-2, -4)\)

\(y = -x^2 -8 x -20 \) ⇒ (\(-4, -4)\)

\(y = 3 x^2 -3 \) ⇒ (\(0, -3)\)

\(y = -x^2 -10 x -22 \) ⇒ (\(-5, 3)\)

\(y = -x^2 -8 x -19 \) ⇒ (\(-4, -3)\)

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

\(y = x^2 + 6 x + 11 \) ⇒ (\(-3, 2)\)

\(y = x^2 + 4 x + 6 \) ⇒ (\(-2, 2)\)

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

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

\(y = -x^2 -12 x -37 \) ⇒ (\(-6, -1)\)

\(y = -x^2 + 8 x -20 \) ⇒ (\(4, -4)\)

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

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

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

\(y = 3 x^2 + 30 x + 71 \) ⇒ (\(-5, -4)\)
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