Consider the quadratic function f(x)=x2−6x+9 . What is the vertex, axis of symmetry, y-intercept, and x-intercepts.
To find the vertex of the quadratic function f(x)=x^2-6x+9, we first need to determine the x-coordinate of the vertex.
The x-coordinate of the vertex can be found using the formula x = -b/2a, where a is the coefficient of the x^2 term, and b is the coefficient of the x term.
In this case, a = 1 and b = -6. So the x-coordinate of the vertex is x = -(-6)/(2*1) = 3.
To find the y-coordinate of the vertex, we substitute the x-coordinate back into the original function: f(3) = 3^2 - 6(3) + 9 = 9 - 18 + 9 = 0.
Therefore, the vertex of the function is (3,0).
The axis of symmetry is the vertical line that passes through the vertex, which in this case is x = 3.
To find the y-intercept, we set x=0 in the equation: f(0)=0^2-6(0)+9=9.
So the y-intercept is 9.
To find the x-intercepts, we set f(x) = 0 and solve for x:
x^2 - 6x + 9 = 0
(x-3)(x-3) = 0
x = 3
Therefore, the x-intercept is 3.