What is the axis y=x^2-13x+40 ?

The axis y=x^2-13x+40 is x= 13/2

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axis y=x^2-13x+40

axis

y = x^{2} - 13 x + 40

x = \frac{13}{2}

Solution steps

y = x^{2} - 13 x + 40

Rewrite

y = x^{2} - 13 x + 40

in the standard form

4 \cdot \frac{1}{4} (y - (- \frac{9}{4})) = (x - \frac{13}{2})^{2}

Therefore parabola properties are:

(h, k) = (\frac{13}{2}, - \frac{9}{4}), 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{13}{2}

y = x^{2} + 2 x - 15

y = - \frac{( x + 3 ) ^{2}}{20} + 1

F = \frac{m v ^{2}}{r}

y^{2} = x - 3

y = 4 (x + 1)^{2} - 3