What is the axis y=2x^2+15x ?

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axis y=2x^2+15x

axis

y = 2 x^{2} + 15 x

x = - \frac{15}{4}

Solution steps

y = 2 x^{2} + 15 x

Rewrite

y = 2 x^{2} + 15 x

in the standard form

4 \cdot \frac{1}{8} (y - (- \frac{225}{8})) = (x - (- \frac{15}{4}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{15}{4}, - \frac{225}{8}), p = \frac{1}{8}

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{15}{4}

y = x^{2} - x - 12

x = (3 y - 1)^{2}

y = \frac{( - 7 x + 3 ) ( 9 x + 4 )}{24}

0 = (y - 1)^{2} - 16 (x + 4)

y = - 1 + 3 x + x^{2}