We are given the function $g(x) = 3x^2 - 2x + 7$, and we are asked to find the values of $g(x)$ for different inputs.

Let's solve each part step by step:

a) $g(0)$

Substitute $x = 0$ into the function: $g(0) = 3(0)^2 - 2(0) + 7 = 0 - 0 + 7 = 7$

b) $g(-2)$

Substitute $x = -2$ into the function: $g(-2) = 3(-2)^2 - 2(-2) + 7 = 3(4) + 4 + 7 = 12 + 4 + 7 = 23$

c) $g(3)$

Substitute $x = 3$ into the function: $g(3) = 3(3)^2 - 2(3) + 7 = 3(9) - 6 + 7 = 27 - 6 + 7 = 28$

d) $g(-x)$

Substitute $x = -x$ into the function: $g(-x) = 3(-x)^2 - 2(-x) + 7 = 3x^2 + 2x + 7$

e) $g(1 - t)$

Substitute $x = 1 - t$ into the function: $g(1 - t) = 3(1 - t)^2 - 2(1 - t) + 7$ First, expand $(1 - t)^2$: $(1 - t)^2 = 1 - 2t + t^2$ Now substitute: $g(1 - t) = 3(1 - 2t + t^2) - 2(1 - t) + 7$ Distribute: $g(1 - t) = 3(1) - 3(2t) + 3(t^2) - 2(1) + 2t + 7$ Simplify: $g(1 - t) = 3 - 6t + 3t^2 - 2 + 2t + 7 = 3t^2 - 4t + 8$

Final answers:

• a) $g(0) = 7$
• b) $g(-2) = 23$
• c) $g(3) = 28$
• d) $g(-x) = 3x^2 + 2x + 7$
• e) $g(1 - t) = 3t^2 - 4t + 8$

Let me know if you'd like further details or have any questions!

Here are some related questions:

1. What is the general formula for finding the vertex of a quadratic function?
2. How would you compute the derivative of $g(x) = 3x^2 - 2x + 7$?
3. Can the function $g(x)$ have any real roots? How would you determine that?
4. How would the graph of $g(x) = 3x^2 - 2x + 7$ look like?
5. What is the value of $g(5)$?

Tip: When expanding squared binomials like $(1 - t)^2$, remember to apply the formula $(a - b)^2 = a^2 - 2ab + b^2$.