To determine if the quadratic function [tex]\( f(x) \)[/tex] and the linear function [tex]\( g(x) = 2x \)[/tex] intersect, we need to analyze their relationship and solve for their intersection points.
### Step-by-Step Solution:
1. Define the functions:
- [tex]\( g(x) = 2x \)[/tex]
- [tex]\( f(x) = ax^2 + bx + c \)[/tex]
2. Set up the equation for intersection points:
- For [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] to intersect, we must have [tex]\( f(x) = g(x) \)[/tex].
- Therefore, [tex]\( ax^2 + bx + c = 2x \)[/tex].
- Rearrange into standard quadratic form: [tex]\( ax^2 + (b-2)x + c = 0 \)[/tex].
3. Analyze the quadratic equation:
- The quadratic equation [tex]\( ax^2 + (b-2)x + c = 0 \)[/tex] has a standard form [tex]\( Ax^2 + Bx + C = 0 \)[/tex] with:
- [tex]\( A = a \)[/tex]
- [tex]\( B = b - 2 \)[/tex]
- [tex]\( C = c \)[/tex]
4. Compute the discriminant:
- The discriminant [tex]\( \Delta \)[/tex] of a quadratic equation [tex]\( Ax^2 + Bx + C = 0 \)[/tex] is given by [tex]\( \Delta = B^2 - 4AC \)[/tex].
- Substitute [tex]\( B = b - 2 \)[/tex], [tex]\( A = a \)[/tex], and [tex]\( C = c \)[/tex]:
[tex]\[ \Delta = (b - 2)^2 - 4ac \][/tex]
5. Evaluate the discriminant:
- The discriminant indicates the nature of the roots:
- If [tex]\( \Delta > 0 \)[/tex], there are two distinct real roots (intersection points).
- If [tex]\( \Delta = 0 \)[/tex], there is exactly one real root (tangent point).
- If [tex]\( \Delta < 0 \)[/tex], there are no real roots (no intersection).
6. Example coefficients:
- Assume example values for coefficients:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = -3 \)[/tex]
- [tex]\( c = 2 \)[/tex]
- Calculate the discriminant:
[tex]\[ \Delta = (-3 - 2)^2 - 4 \cdot 1 \cdot 2 = (-5)^2 - 8 = 25 - 8 = 17 \][/tex]
7. Interpret the result:
- Since [tex]\( \Delta = 17 \)[/tex] which is greater than 0, the quadratic function [tex]\( f(x) \)[/tex] and the linear function [tex]\( g(x) = 2x \)[/tex] have two distinct real intersection points.
### Conclusion:
The quadratic function [tex]\( f(x) \)[/tex] and the linear function [tex]\( g(x) = 2x \)[/tex] will intersect at both negative and positive [tex]\( x \)[/tex]-coordinates. Thus, the correct answer is:
Yes, at negative and positive [tex]\( x \)[/tex]-coordinates.
