Answer :
Let's analyze each statement one by one for a quadratic function of the form [tex]\( f(x) = a x^2 + bx + c \)[/tex] when [tex]\( b = 0 \)[/tex], simplifying the function to [tex]\( f(x) = ax^2 + c \)[/tex].
1. The graph will always have zero [tex]\( x \)[/tex]-intercepts.
- The [tex]\( x \)[/tex]-intercepts occur where the function [tex]\( f(x) = ax^2 + c \)[/tex] equals 0, i.e., [tex]\( ax^2 + c = 0 \)[/tex].
- Solving for [tex]\( x \)[/tex], we get [tex]\( x^2 = -\frac{c}{a} \)[/tex].
- The number of solutions depends on the values of [tex]\( a \)[/tex] and [tex]\( c \)[/tex]:
- If [tex]\( c = 0 \)[/tex], there is one solution: [tex]\( x = 0 \)[/tex].
- If [tex]\( a \)[/tex] and [tex]\( c \)[/tex] are both non-zero and have opposite signs (one positive and one negative), there are two real solutions.
- If [tex]\( a \)[/tex] and [tex]\( c \)[/tex] have the same signs (both positive or both negative), there are no real solutions.
- Therefore, the statement is false because the graph can have zero, one, or two [tex]\( x \)[/tex]-intercepts.
2. The function will always have a minimum.
- For [tex]\( f(x) = ax^2 + c \)[/tex], the direction the parabola opens depends on the sign of [tex]\( a \)[/tex]:
- If [tex]\( a > 0 \)[/tex], the parabola opens upwards, having a minimum at the vertex.
- If [tex]\( a < 0 \)[/tex], the parabola opens downwards, having a maximum at the vertex.
- Therefore, the statement is true if [tex]\( a > 0 \)[/tex], but since there's no restriction on [tex]\( a \)[/tex] always being greater than zero, it is more accurate to say it sometimes has a minimum.
3. The [tex]\( y \)[/tex]-intercept will always be the vertex.
- The [tex]\( y \)[/tex]-intercept is the point where [tex]\( x = 0 \)[/tex], which gives [tex]\( f(0) = c \)[/tex].
- The vertex of the parabola [tex]\( ax^2 + c \)[/tex] when [tex]\( b = 0 \)[/tex] is at [tex]\( (0, c) \)[/tex].
- Therefore, the [tex]\( y \)[/tex]-intercept (0, c) is indeed the vertex.
- The statement is true.
4. The axis of symmetry will always be positive.
- The axis of symmetry for a quadratic function [tex]\( f(x) = ax^2 + bx + c \)[/tex] is given by [tex]\( x = -\frac{b}{2a} \)[/tex].
- When [tex]\( b = 0 \)[/tex], this reduces to [tex]\( x = 0 \)[/tex].
- [tex]\( x = 0 \)[/tex] is neither positive nor negative, so the statement is false.
In summary, the true statements are:
- The [tex]\( y \)[/tex]-intercept will always be the vertex.
- The function will always have a minimum.
Thus, the correct choices are:
- The function will always have a minimum.
- The [tex]\( y \)[/tex]-intercept will always be the vertex.
Therefore, the answer is:
[tex]\[ (2, 3) \][/tex]
1. The graph will always have zero [tex]\( x \)[/tex]-intercepts.
- The [tex]\( x \)[/tex]-intercepts occur where the function [tex]\( f(x) = ax^2 + c \)[/tex] equals 0, i.e., [tex]\( ax^2 + c = 0 \)[/tex].
- Solving for [tex]\( x \)[/tex], we get [tex]\( x^2 = -\frac{c}{a} \)[/tex].
- The number of solutions depends on the values of [tex]\( a \)[/tex] and [tex]\( c \)[/tex]:
- If [tex]\( c = 0 \)[/tex], there is one solution: [tex]\( x = 0 \)[/tex].
- If [tex]\( a \)[/tex] and [tex]\( c \)[/tex] are both non-zero and have opposite signs (one positive and one negative), there are two real solutions.
- If [tex]\( a \)[/tex] and [tex]\( c \)[/tex] have the same signs (both positive or both negative), there are no real solutions.
- Therefore, the statement is false because the graph can have zero, one, or two [tex]\( x \)[/tex]-intercepts.
2. The function will always have a minimum.
- For [tex]\( f(x) = ax^2 + c \)[/tex], the direction the parabola opens depends on the sign of [tex]\( a \)[/tex]:
- If [tex]\( a > 0 \)[/tex], the parabola opens upwards, having a minimum at the vertex.
- If [tex]\( a < 0 \)[/tex], the parabola opens downwards, having a maximum at the vertex.
- Therefore, the statement is true if [tex]\( a > 0 \)[/tex], but since there's no restriction on [tex]\( a \)[/tex] always being greater than zero, it is more accurate to say it sometimes has a minimum.
3. The [tex]\( y \)[/tex]-intercept will always be the vertex.
- The [tex]\( y \)[/tex]-intercept is the point where [tex]\( x = 0 \)[/tex], which gives [tex]\( f(0) = c \)[/tex].
- The vertex of the parabola [tex]\( ax^2 + c \)[/tex] when [tex]\( b = 0 \)[/tex] is at [tex]\( (0, c) \)[/tex].
- Therefore, the [tex]\( y \)[/tex]-intercept (0, c) is indeed the vertex.
- The statement is true.
4. The axis of symmetry will always be positive.
- The axis of symmetry for a quadratic function [tex]\( f(x) = ax^2 + bx + c \)[/tex] is given by [tex]\( x = -\frac{b}{2a} \)[/tex].
- When [tex]\( b = 0 \)[/tex], this reduces to [tex]\( x = 0 \)[/tex].
- [tex]\( x = 0 \)[/tex] is neither positive nor negative, so the statement is false.
In summary, the true statements are:
- The [tex]\( y \)[/tex]-intercept will always be the vertex.
- The function will always have a minimum.
Thus, the correct choices are:
- The function will always have a minimum.
- The [tex]\( y \)[/tex]-intercept will always be the vertex.
Therefore, the answer is:
[tex]\[ (2, 3) \][/tex]
