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:

Step1: Analyze each option

• Option 1: When \(b = 0\), the function is \(f(x)=ax^{2}+c\). The number of \(x\)-intercepts depends on the discriminant \(\Delta=b^{2}-4ac = - 4ac\). If \(ac<0\), there are two \(x\)-intercepts, so this option is false.
• Option 2: The direction of the parabola (whether it has a minimum or maximum) depends on the sign of \(a\). If \(a<0\), the function has a maximum, not a minimum, so this option is false.
• Option 3: The \(y\)-intercept of \(f(x)=ax^{2}+bx + c\) is at \(x = 0\), so \(f(0)=c\). The vertex of a parabola \(f(x)=ax^{2}+bx + c\) has \(x\)-coordinate \(-\frac{b}{2a}\). When \(b = 0\), the \(x\)-coordinate of the vertex is \(0\), and \(f(0)=c\), so the \(y\)-intercept (which is \((0,c)\)) is the vertex. This option is true.
• Option 4: The axis of symmetry is \(x=-\frac{b}{2a}\). When \(b = 0\), the axis of symmetry is \(x = 0\) (the \(y\)-axis), which is not positive (except when considering the non - negative aspect, but \(x = 0\) is not positive in the sense of being greater than \(0\)). So this option is false.

Answer:

The \(y\)-intercept will always be the vertex. (The option corresponding to this statement among the given options)