What is the axis y=4x^2+x+1 ?

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axis y=4x^2+x+1

axis

y = 4 x^{2} + x + 1

x = - \frac{1}{8}

Solution steps

y = 4 x^{2} + x + 1

Rewrite

y = 4 x^{2} + x + 1

in the standard form

4 \cdot \frac{1}{16} (y - \frac{15}{16}) = (x - (- \frac{1}{8}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{1}{8}, \frac{15}{16}), 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 = - \frac{1}{8}

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

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

y = 18 x^{2} - 14 x + 9

y = - (- 12 x + 13)^{2} - 1

y = 3 x^{2} - 5