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

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Answer 1 · 2018-01-03T17:15:42

Vertex $(3/2, 23/4)$
Axis of symmetry: $x=3/2$

Given a quadratic of the form $y=ax^2+bx+c$ the vertex, $(h,k)$ is of the form $h=-b/(2a)$ and $k$ is found by substituting $h$ .

$y=x^2-3x+8$ gives $h=-(-3)/(2*1)=3/2$ .

To find $k$ we substitute this value back in:

$k=(3/2)^2-3(3/2)+8 = 9/4-9/2+8 = 23/4$ .

So the vertex is $(3/2, 23/4)$ .

The axis of symmetry is the vertical line through the vertex, so in this case it is $x=3/2$ .

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