What is the axis y=x^2-7x+12 ?

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axis y=x^2-7x+12

axis

y = x^{2} - 7 x + 12

x = \frac{7}{2}

Solution steps

y = x^{2} - 7 x + 12

Rewrite

y = x^{2} - 7 x + 12

in the standard form

4 \cdot \frac{1}{4} (y - (- \frac{1}{4})) = (x - \frac{7}{2})^{2}

Therefore parabola properties are:

(h, k) = (\frac{7}{2}, - \frac{1}{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{7}{2}

y \geq x^{2}

y = 5 (x + 2)^{2} - 3

y = - \frac{1}{2} x^{2} - 5 x - \frac{15}{2}

x = y - y^{2}

y = x^{2} - 5 x