What is the axis y=(2x^2+7x+12)/2 ?

The axis y=(2x^2+7x+12)/2 is x=-7/4

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axis y=(2x^2+7x+12)/2

axis

y = \frac{2 x ^{2} + 7 x + 12}{2}

x = - \frac{7}{4}

Solution steps

y = \frac{2 x ^{2} + 7 x + 12}{2}

Rewrite

y = \frac{2 x ^{2} + 7 x + 12}{2}

in the standard form

4 \cdot \frac{1}{4} (y - \frac{47}{16}) = (x - (- \frac{7}{4}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{7}{4}, \frac{47}{16}), p = \frac{1}{4}

Parabola is of the form

4 p (y - k) = (x - h)^{2}

and is symmetric around the

y

-axis

Axis of symmetry is a line parallel to the

y

-axis which intersects the vertex:

x = - \frac{7}{4}

- x^{2} + 4 x - 11

y = 5 x^{2} - 9

4 x^{2} - 24 x - 40 y - 4 = 0

x = + (y + 11)^{2} - 14

y = x^{2} - 3 x + 28