How do you find the axis of symmetry, vertex and x intercepts for #y=-1/3x^2+x-3#?

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Answer 1 · 2016-10-02T16:58:46

axis of symmetry: #x = 3/2#
vertex: #(3/2, -9/4)#
No x intercepts

The axis of symmetry of a parabola described by an equation of the form #y = ax² + bx + c# is a vertical line described by the equation #x = -b/2a#

In this case #a = -1/3# and #b = 1#, therefore, the axis of symmetry is the vertical line described by the equation:

#x = 3/2#

The vertex is a point where the x coordinate is that of the axis of symmetry and the y coordinate is the function at the axis of symmetry:

#y(3/2) = -1/3(3/2)² + 3/2 - 3#

#y(3/2) = -3/4 + 6/4 - 12/4#

#y(3/2) = -9/4#

The vertex is #(3/2, -9/4)#

Let's check the discriminant of the quadratic:

#b² - 4(a)(c) = 1² - (4)(-1/3)(-3)#

#b² - 4(a)(c) = -3#

There are no real roots, therefore the function does not cross the x axis.