What is the axis y=+(8-x)(3x+2) ?

The axis y=+(8-x)(3x+2) is x= 11/3

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axis y=+(8-x)(3x+2)

axis

y = + (8 - x) (3 x + 2)

x = \frac{11}{3}

Solution steps

y = (8 - x) (3 x + 2)

Rewrite

y = (8 - x) (3 x + 2)

in the standard form

4 (- \frac{1}{12}) (y - \frac{169}{3}) = (x - \frac{11}{3})^{2}

Therefore parabola properties are:

(h, k) = (\frac{11}{3}, \frac{169}{3}), p = - \frac{1}{12}

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{11}{3}

y = 4 + 2 x - x^{2}

y = - \frac{2}{5} x^{2} + 4 x

y = - x^{2} - 6 \cdot x + 7

y = x^{2} - 3 x + 5

y = \frac{3}{8} x^{2}