The domain of y = ax^2 + c is all real numbers. Describe the range when (a) a > 0) and (b) a < 0. a. range when a > 0 b. range when a < 0

[tex]\begin{gathered} y=ax^2+c \\ \downarrow\text{substracting c both sides} \\ y-c=ax^2 \\ \downarrow\text{ dividing by a both sides} \\ \frac{y-c}{a}=x^2 \\ \downarrow\text{ square root of both sides} \\ \sqrt{\frac{y-c}{a}}=x \end{gathered}[/tex]

We know that in the real numbers, the square root of a negative number doesn't exist. Then

[tex]\begin{gathered} \sqrt[]{\frac{y-c}{a}}=x \\ \downarrow \\ \frac{y-c}{a}\text{ cannot be negative} \\ \downarrow\frac{y-c}{a}\text{ is positive or 0} \\ \frac{y-c}{a}\ge0 \end{gathered}[/tex]

Step 2

Finding the range

When a > 0

Then

[tex]\begin{gathered} \frac{y-c}{a}is\text{ positive if and only if } \\ y-c\ge0 \end{gathered}[/tex]

Adding c both sides:

[tex]\begin{gathered} y-c\ge0 \\ y\ge c \end{gathered}[/tex]

Then, y goes from c to infinity

[tex]y\in\lbrack c,\infty)[/tex]

Answer A - Range = [c, ∞) when a >0

When a < 0

Then

[tex]\begin{gathered} \frac{y-c}{a}is\text{ positive if and only if } \\ y-c\leq0 \end{gathered}[/tex]

Because the division of two negative numbers is always positive

Adding c both sides:

[tex]\begin{gathered} y-c\leq0 \\ y\leq c \end{gathered}[/tex]

Then, y goes from minus infinity to c

[tex]y\in(-\infty,c\rbrack[/tex]

Answer - Range = (- ∞, c] when a < 0