Let's go through each part step-by-step:

Given the function:
$h(x) = x^2 + 1$

(a) Find $h(4)$:

Substitute $x = 4$ into the function:

$h(4) = 4^2 + 1 = 16 + 1 = 17$

(b) Find $h(-3)$:

Substitute $x = -3$ into the function:

$h(-3) = (-3)^2 + 1 = 9 + 1 = 10$

(c) If $h(t) = 26$, find the possible values of $t$:

We are given that $h(t) = 26$, which means:

$t^2 + 1 = 26$

Now solve for $t$:

$t^2 = 26 - 1 = 25$

Taking the square root of both sides:

$t = \pm \sqrt{25} = \pm 5$

So, the possible values of $t$ are $t = 5$ or $t = -5$.

(d) Find $h(5.75)$:

Substitute $x = 5.75$ into the function:

$h(5.75) = (5.75)^2 + 1$

Now calculate $5.75^2$:

$5.75^2 = 33.0625$

Thus:

$h(5.75) = 33.0625 + 1 = 34.0625$

Summary of Answers:

• (a) $h(4) = 17$
• (b) $h(-3) = 10$
• (c) $t = 5$ or $t = -5$
• (d) $h(5.75) = 34.0625$

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Do you want details on any step? Here are 5 related questions:

1. How would you describe the graph of $h(x) = x^2 + 1$?
2. What are the intercepts of the function $h(x) = x^2 + 1$?
3. Can you find $h(0)$?
4. What is the vertex of the function $h(x) = x^2 + 1$?
5. What is the range of the function $h(x) = x^2 + 1$?

Tip: When solving quadratic equations like $t^2 = 25$, always remember to consider both positive and negative roots!