What is the axis y=+x^2+3x-4 ?

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axis y=+x^2+3x-4

axis

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

x = - \frac{3}{2}

Solution steps

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

Rewrite

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

in the standard form

4 \cdot \frac{1}{4} (y - (- \frac{25}{4})) = (x - (- \frac{3}{2}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{3}{2}, - \frac{25}{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{3}{2}

y = x^{2} + 8 x + 15

y = (2 x - 1) (5 x - 6)

y = 1 - x - x^{2}

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

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