What is the axis y=x^2-42+10x ?

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axis y=x^2-42+10x

axis

y = x^{2} - 42 + 10 x

x = - 5

Solution steps

y = x^{2} - 42 + 10 x

Rewrite

y = x^{2} - 42 + 10 x

in the standard form

4 \cdot \frac{1}{4} (y - (- 67)) = (x - (- 5))^{2}

Therefore parabola properties are:

(h, k) = (- 5, - 67), 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 = - 5

(x + 9)^{2} = - 44 (y - 1)

y = - 2 x^{2} + 12 x - 14

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

y = - 6 (x - 5)^{2} + 12

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