What is the axis y=4x^2-40x+98 ?

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axis y=4x^2-40x+98

axis

y = 4 x^{2} - 40 x + 98

x = 5

Solution steps

y = 4 x^{2} - 40 x + 98

Rewrite

y = 4 x^{2} - 40 x + 98

in the standard form

4 \cdot \frac{1}{16} (y - (- 2)) = (x - 5)^{2}

Therefore parabola properties are:

(h, k) = (5, - 2), p = \frac{1}{16}

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 = 5

y = x^{2} + x - 24

y = \frac{x ^{2}}{2} + 6 x + 21

y = x^{2} - 10000 x - 24000000

y = 6 (x - \frac{1}{2}) (x + 3)

y = 9 x^{2} + 25