What is the axis y=4x^2-8x+7 ?

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axis y=4x^2-8x+7

axis

y = 4 x^{2} - 8 x + 7

x = 1

Solution steps

y = 4 x^{2} - 8 x + 7

Rewrite

y = 4 x^{2} - 8 x + 7

in the standard form

4 \cdot \frac{1}{16} (y - 3) = (x - 1)^{2}

Therefore parabola properties are:

(h, k) = (1, 3), 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 = 1

x^{2} = - 6 y

+ 5 x^{2} + 30 x + 24 y = 51

x^{2} = - 4 y

y = - 5 x^{2} + 118 x + 7

y = \frac{6 x}{7} \cdot \frac{4 x}{3}