Answer:
See explanation
Step-by-step explanation:
The question is not properly presented; I will answer this with the following similar question.
[tex]h(x) = -\frac{1}{3}x^2 + 4[/tex] --- [tex]x \ne 2[/tex]
[tex]h(x) = -4[/tex] --- [tex]x= 2[/tex]
Required
[tex]h(-5)\ \&\ h(2)[/tex]
Calculating h(-5)
This implies that: x = -5
[tex]-5 \ne 2[/tex]
So, we make use of:
[tex]h(x) = -\frac{1}{3}x^2 + 4[/tex]
Substitute -5 for x
[tex]h(-5) = -\frac{1}{3} * (-5)^2 + 4[/tex]
[tex]h(-5) = -\frac{1}{3} * 25 + 4[/tex]
[tex]h(-5) = -\frac{25}{3} + 4[/tex]
Take LCM
[tex]h(-5) = \frac{-25+12}{3}[/tex]
[tex]h(-5) = -\frac{13}{3}[/tex]
Calculating h(2)
This implies that: x = 2
[tex]2 = 2[/tex]
So, we make use of:
[tex]h(x) = -4[/tex]
Substitute 2 for x
[tex]h(2) = -4[/tex]
Use the above explanation to answer your question
