How many intercepts does #y = x^2 − 6x + 9# have?

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Answer 1 · 2015-08-12T23:51:10

One #x#-intercept and one #y#-intercept.

You can determine #x#-intercepts by making the function equal to zero and #y#-intercepts by evaluating the function for #x=0#.

When #y=0#, you have

#x^2 - 6x + 9 = 0#

In order to determine how many solutions this quadratic equation has, you can calculate the value of its discriminant, #Delta#.

For a quadratic equation that takes the general form

#color(blue)(ax^2 + bx + c = 0)#

the discriminant is equal to

#color(blue)(Delta = b^2 - 4ac)#

In your case, you have #a=1#, #b=-6#, and #c=9#, which means that the discriminant is equal to

#Delta = (-6)^2 - 4 * 1 * 9#

#Delta = 36 - 36 = color(green)(0)#

When the discriminant is equal to zero, your equation will only have one real solution (a repeated root) that takes the form

#x = (-b +- sqrt(Delta))/(2a) = (-b +- 0)/(2a) = -b/(2a)#

In your case, the root will be

#x = -((-6))/(2 * 1) = 6/2 = 3#

This means that the function has one #x#-intercept, #x=3#.

The #y#-intercept will be

#y = (0)^2 - 6 * (0) + 9 = 9#

The function will thus intercept the #y#-axis in the point #(0,9)# and the #x#-acis in the point #(3,0)#.

graph{x^2 - 6x + 9 [-10, 10, -5, 5]}