What is the axis of symmetry and vertex for the graph #y = 3x^2 - 7x - 8#?

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Answer 1 · 2017-05-04T22:47:02

The axis of symmetry is #x=7/6# and the vertex #(7/6, -145/12)#

Given a quadratic equation representing a parabola in the form:

#y = ax^2+bx+c#

we can convert to vertex form by completing the square:

#y = ax^2+bx+c#

#color(white)(y) = a(x-(-b)/(2a))^2+(c-b^2/(4a))#

#color(white)(y) = a(x-h)^2+k#

with vertex #(h, k) = (-b/(2a), c-b^2/(4a))#.

The axis of symmetry is the vertical line #x=-b/(2a)#.

In the given example, we have:

#y = 3x^2-7x-8#

#color(white)(y) = 3(x-7/6)^2-(8+49/12)#

#color(white)(y) = 3(x-7/6)^2-145/12#

So the axis of symmetry is #x=7/6# and the vertex #(7/6, -145/12)#

graph{(y-(3x^2-7x-8))(4(x-7/6)^2+(y+145/12)^2-0.01)(x-7/6) = 0 [-5.1, 5.1, -13.2, 1.2]}