Answer: This is a quadratic function in standard form:
f(x) = x^2 - 10x + 9
The vertex form of a quadratic function is given by:
f(x) = a(x - h)^2 + k
where (h, k) is the vertex of the parabola.
To rewrite the given function in vertex form, we need to complete the square:
f(x) = x^2 - 10x + 9
= (x - 5)^2 - 16
Now we can see that the vertex is at (5, -16) and the axis of symmetry is x = 5.
To find the x-intercepts, we set y = 0:
0 = (x - 5)^2 - 16
16 = (x - 5)^2
±4 = x - 5
x = 1 or x = 9
Therefore, the x-intercepts are (1, 0) and (9, 0).
To find the y-intercept, we set x = 0:
f(0) = 9
Therefore, the y-intercept is (0, 9).
The graph of the function is a parabola that opens upward with vertex at (5, -16), x-intercepts at (1, 0) and (9, 0), and y-intercept at (0, 9).
Step-by-step explanation:
the vertex is at (5, -16)
y-intercept is (0, 9).
the x-intercepts are (1, 0) and (9, 0).
axis of symmetry is x = 5
the range of the function is all real numbers greater than or equal to -16. In interval notation, we can write this as [-16, ∞).
the domain of the given function f(x) = x^2 - 10x + 9 is all real numbers.