What is the axis y=x^2+x-10 ?

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axis y=x^2+x-10

axis

y = x^{2} + x - 10

x = - \frac{1}{2}

Solution steps

y = x^{2} + x - 10

Rewrite

y = x^{2} + x - 10

in the standard form

4 \cdot \frac{1}{4} (y - (- \frac{41}{4})) = (x - (- \frac{1}{2}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{1}{2}, - \frac{41}{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{1}{2}

y = x^{2} - 8 x + 19

y = x^{2} + 4 x + 3

x^{2} + 6 x + 8 y = 7

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

y = 0.5 x^{2} - 3 x + 5