Question

for all functions of the form $f(x) = ax^2 + bx + c$, which is true when $b = 0$?
the graph will always have zero $x$-intercepts.
the function will always have a minimum.
the $y$-intercept will always be the vertex.
the axis of symmetry will always be positive.

Explanation:

1. Analyze the first option: When \( b = 0 \), the function is \( f(x)=ax^{2}+c \). The number of \( x \)-intercepts depends on \( a \) and \( c \). For example, if \( a = 1 \) and \( c=- 1 \), \( f(x)=x^{2}-1=(x - 1)(x + 1) \) has two \( x \)-intercepts. So the first option is false.
2. Analyze the second option: The function \( f(x)=ax^{2}+c \) has a minimum if \( a>0 \) and a maximum if \( a < 0 \). For example, if \( a=-1 \) and \( c = 0 \), \( f(x)=-x^{2} \) has a maximum at \( x = 0 \). So the second option is false.
3. Analyze the third option: The \( y \)-intercept of \( f(x)=ax^{2}+bx + c \) is found by setting \( x = 0 \), so \( f(0)=c \). The vertex of a quadratic function \( f(x)=ax^{2}+bx + c \) has its \( x \)-coordinate given by \( x=-\frac{b}{2a} \). When \( b = 0 \), the \( x \)-coordinate of the vertex is \( x = 0 \). Then \( f(0)=c \), which is the \( y \)-intercept. So the \( y \)-intercept is the vertex when \( b = 0 \).
4. Analyze the fourth option: The axis of symmetry of \( f(x)=ax^{2}+bx + c \) is \( x=-\frac{b}{2a} \). When \( b = 0 \), the axis of symmetry is \( x = 0 \) (the \( y \)-axis), which is not positive. So the fourth option is false.

Answer:

The y - intercept will always be the vertex.