Answer:
Hello,
Step-by-step explanation:
y-intercept: x=0 ,f(0)=-10
[tex]x^3-8x^2+17x-10\\\\=x^3-5x^2-3x^2+15x+2x-10\\\\=x^2(x-5)-3x(x-5)+2(x-5)\\\\=(x-5)(x^2-3x+2)\\\\=(x-5)(x^2-x-2x+2)\\\\=(x-5)(x(x-1)-2(x-1))\\\\=(x-5)(x-1)(x-2)\\[/tex]
x-intercepts are: 5,1,2.
Dividing the polynomials and solving a quadratic equation, we find that:
The function is:
[tex]f(x) = x^3 - 8x^2 + 17x - 10[/tex]
One factor is:
[tex]x - 5[/tex]
This means that:
[tex](x - 5)(ax^2 + bx + c) = x^3 - 8x^2 + 17x - 10[/tex]
Then
[tex]ax^3 + (b - 5a)x^2 + (c - 5b)x - 5c = x^3 - 8x^2 + 17x - 10[/tex]
Equaling both sides, we have that:
[tex]a = 1[/tex]
[tex]-5c = -10 \rightarrow c = 2[/tex]
[tex]b - 5a = -8[/tex]
[tex]b = -3[/tex]
Thus, the other two x-intercepts are the roots of:
[tex]x^2 - 3x + 2 = 0[/tex]
Which is a quadratic equation with [tex]a = 1, b = -3, c = 2[/tex].
Applying Bhaskara:
[tex]\Delta = (-3)^2 - 4(1)(2) = 1[/tex]
[tex]x_{1} = \frac{-(-3) + \sqrt{1}}{2(1)} = 2[/tex]
[tex]x_{2} = \frac{-(-3) - \sqrt{1}}{2(1)} = 1[/tex]
Thus, the other two x-intercepts are: [tex]x = 2, x = 1[/tex].
The y-intercept is f(0), thus:
[tex]y = f(0) = 0^3 - 8(0)^2 + 17(0) - 10 = -10[/tex]
The y-intercept is [tex]y = -10[/tex].
A similar problem is given at https://brainly.com/question/24662212
