What is the axis of symmetry and vertex for the graph y = 9x^2 - 27x + 20?

1 Answer
Jul 13, 2018

The axis of symmetry is x=3/2.

The vertex is (3/2,-1/4).

Explanation:

Given:

y=9x^2-27x+20 is a quadratic equation in standard form:

y=ax^2+bx+c,

where:

a=9, b=027, c=20

The formula for the axis of symmetry is:

x=(-b)/(2a)

x=(-(-27))/(2*9)

x=27/18

Reduce by dividing the numerator and denominator by 9.

x=(27-:9)/(18-:9)

x=3/2

The axis of symmetry is x=3/2. This is also the x-coordinate of the vertex.

To find the y-coordinate of the vertex, substitute 3/2 for x in the equation and solve for y.

y=9(3/2)^2-27(3/2)+20

y=9(9/4)-81/2+20

y=81/4-81/2+20

The least common denominator is 4. Multiply 81/2 by 2/2 and 20 by 4/4 to get equivalent fractions with 4 as the denominator. Since n/n=1, the numbers will change but the value of the fractions will remain the same.

y=81/4-(81/2xx2/2)+(20xx4/4)

y=81/4-162/4+80/4

y=(81-162+80)/4

y=-1/4

The vertex is (3/2,-1/4).

graph{y=9x^2-27x+20 [-10, 10, -5, 5]}