for all functions of the form f(x) = ax² + bx + c, which is true when b = 0? the graph will always have zero…

Explanation:

Step1: Recall the properties of quadratic functions

The general form of a quadratic function is $f(x)=ax^{2}+bx + c$. The axis - of - symmetry is given by the formula $x =-\frac{b}{2a}$, the $y$-intercept is $f(0)=c$, and the vertex has $x$-coordinate $x =-\frac{b}{2a}$. When $b = 0$, the function becomes $f(x)=ax^{2}+c$.

Step2: Analyze the $x$-intercepts

Set $y = 0$, so $ax^{2}+c=0$, or $x^{2}=-\frac{c}{a}$. If $a$ and $c$ have the same sign, there are no real $x$-intercepts. For example, if $a = 1$ and $c = 1$, then $x^{2}=- 1$ has no real solutions. So the graph does not always have zero $x$-intercepts.

Step3: Analyze the minimum or maximum

The function $y = ax^{2}+c$ has a minimum if $a>0$ and a maximum if $a < 0$. So it does not always have a minimum.

Step4: Analyze the $y$-intercept and vertex

The $y$-intercept is found by setting $x = 0$, so $y=f(0)=c$. The $x$-coordinate of the vertex of $y=ax^{2}+bx + c$ is $x=-\frac{b}{2a}$. When $b = 0$, the $x$-coordinate of the vertex is $x = 0$. Substituting $x = 0$ into $y=ax^{2}+c$ gives $y = c$. So the $y$-intercept ($x = 0,y = c$) is the vertex.

Step5: Analyze the axis of symmetry

The axis of symmetry of $y=ax^{2}+bx + c$ is $x=-\frac{b}{2a}$. When $b = 0$, the axis of symmetry is $x = 0$ (the $y$-axis), which is not positive.

Answer:

The y-intercept will always be the vertex.