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Mathematics → Precalculus → Quadratic Equations → Finding Vertex of Parabola

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

Questions


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

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

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

\(y = 3 x^2 -18 x + 23\)

\(y = 2 x^2 + 24 x + 74\)

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

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

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

\(y = x^2 + 12 x + 35\)

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

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

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

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

\(y = -2 x^2 + 20 x -49\)

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

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

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

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

\(y = 3 x^2 + 24 x + 50\)

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

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

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

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

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

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

\(y = 2 x^2 + 16 x + 28\)

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

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

Answers


\(y = 3 x^2 + 36 x + 111 \) ⇒ (\(-6, 3)\)
\(y = x^2 -12 x + 39 \) ⇒ (\(6, 3)\)
\(y = -2 x^2 + 20 x -52 \) ⇒ (\(5, -2)\)
\(y = 3 x^2 -18 x + 23 \) ⇒ (\(3, -4)\)

\(y = 2 x^2 + 24 x + 74 \) ⇒ (\(-6, 2)\)
\(y = x^2 + 2 x \) ⇒ (\(-1, -1)\)
\(y = 3 x^2 -30 x + 73 \) ⇒ (\(5, -2)\)
\(y = -x^2 -4 x -3 \) ⇒ (\(-2, 1)\)

\(y = x^2 + 12 x + 35 \) ⇒ (\(-6, -1)\)
\(y = -2 x^2 -4 \) ⇒ (\(0, -4)\)
\(y = x^2 -12 x + 32 \) ⇒ (\(6, -4)\)
\(y = x^2 + 8 x + 14 \) ⇒ (\(-4, -2)\)

\(y = -x^2 -3 \) ⇒ (\(0, -3)\)
\(y = -2 x^2 + 20 x -49 \) ⇒ (\(5, 1)\)
\(y = 3 x^2 + 12 x + 12 \) ⇒ (\(-2, 0)\)
\(y = x^2 -4 x + 6 \) ⇒ (\(2, 2)\)

\(y = -x^2 -8 x -15 \) ⇒ (\(-4, 1)\)
\(y = -2 x^2 -12 x -21 \) ⇒ (\(-3, -3)\)
\(y = 3 x^2 + 24 x + 50 \) ⇒ (\(-4, 2)\)
\(y = 2 x^2 -4 x -2 \) ⇒ (\(1, -4)\)

\(y = x^2 + 8 x + 13 \) ⇒ (\(-4, -3)\)
\(y = -x^2 + 10 x -26 \) ⇒ (\(5, -1)\)
\(y = x^2 + 2 x + 1 \) ⇒ (\(-1, 0)\)
\(y = -x^2 + 12 x -39 \) ⇒ (\(6, -3)\)

\(y = x^2 + 4 x + 4 \) ⇒ (\(-2, 0)\)
\(y = 2 x^2 + 16 x + 28 \) ⇒ (\(-4, -4)\)
\(y = -x^2 -12 x -39 \) ⇒ (\(-6, -3)\)
\(y = 3 x^2 -6 x + 3 \) ⇒ (\(1, 0)\)
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