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MathematicsPrecalculusQuadratic Equations → Finding Vertex of Parabola

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

Questions


\(y = x^2 -10 x + 26\)

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

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

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

\(y = -2 x^2 -16 x -34\)

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

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

\(y = -x^2 + 12 x -39\)

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

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

\(y = 2 x^2 -12 x + 19\)

\(y = -x^2 -10 x -26\)

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

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

\(y = -2 x^2 + 20 x -50\)

\(y = 2 x^2\)

\(y = -2 x^2 + 24 x -76\)

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

\(y = -2 x^2 + 24 x -70\)

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

\(y = 2 x^2 -20 x + 48\)

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

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

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

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

\(y = -2 x^2 + 16 x -29\)

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

\(y = -2 x^2\)

Answers


\(y = x^2 -10 x + 26 \) ⇒ (\(5, 1)\)

\(y = -2 x^2 + 2 \) ⇒ (\(0, 2)\)

\(y = -x^2 + 8 x -15 \) ⇒ (\(4, 1)\)

\(y = -x^2 + 8 x -19 \) ⇒ (\(4, -3)\)

\(y = -2 x^2 -16 x -34 \) ⇒ (\(-4, -2)\)

\(y = x^2 -12 x + 38 \) ⇒ (\(6, 2)\)

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

\(y = -x^2 + 12 x -39 \) ⇒ (\(6, -3)\)

\(y = x^2 -2 x + 1 \) ⇒ (\(1, 0)\)

\(y = -x^2 -6 x -6 \) ⇒ (\(-3, 3)\)

\(y = 2 x^2 -12 x + 19 \) ⇒ (\(3, 1)\)

\(y = -x^2 -10 x -26 \) ⇒ (\(-5, -1)\)

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

\(y = x^2 -10 x + 25 \) ⇒ (\(5, 0)\)

\(y = -2 x^2 + 20 x -50 \) ⇒ (\(5, 0)\)

\(y = 2 x^2 \) ⇒ (\(0, 0)\)

\(y = -2 x^2 + 24 x -76 \) ⇒ (\(6, -4)\)

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

\(y = -2 x^2 + 24 x -70 \) ⇒ (\(6, 2)\)

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

\(y = 2 x^2 -20 x + 48 \) ⇒ (\(5, -2)\)

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

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

\(y = -x^2 + 2 x \) ⇒ (\(1, 1)\)

\(y = -x^2 + 2 x + 1 \) ⇒ (\(1, 2)\)

\(y = -2 x^2 + 16 x -29 \) ⇒ (\(4, 3)\)

\(y = 2 x^2 + 2 \) ⇒ (\(0, 2)\)

\(y = -2 x^2 \) ⇒ (\(0, 0)\)
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