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Sagot :
Range of a function
We know that the range of a function corresponds to the y-values it takes. For the function:
[tex]y=ax^2+c[/tex]we want to find the y values it cannot take.
Step 1
Solving the equation for x.
We want to rearrange the equation:
[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
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