Given the quadratic equation:
![g(x)=3x^2-5x+4](https://img.qammunity.org/2023/formulas/mathematics/college/me4l1kgimenvrfqo03yaqhjlrx8cxvdc78.png)
Let's solve for the following:
• (a) g(-3)
To solve for g(-3), substitute -3 for x and evaluate.
Thus, we have:
![\begin{gathered} g(x)=3x^2-5x+4 \\ \\ g(-3)=3(-3)^2-5(-3)+4 \\ \\ g(-3)=3(9)+15+4 \\ \\ g(-3)=27+15+4 \\ \\ g(-3)=46 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/5yladx2zh2d5zlktij6f1wnoqlesgo53n7.png)
Hence, we have:
g(-3) = 46
• (b) To determine the coordinates of the point given in question (a).
In the function, g(x) can also be written as y.
Thus, from g(-3), we have the following:
x = -3
y = 46
When x = -3, the value of y = 46
In point form, we have the coordinates:
(x, y) ==> (-3, 46)
Therefore, the coordinates of the given point by the answer in (a) is:
(-3, 46)
• (c) Evaluate g(2a).
To evaluate g(2a), substitute 2a for x in the equation and evaluate.
Thus, we have:
![\begin{gathered} g(x)=3x^2-5x+4 \\ \\ g(2a)=3(2a)^2-5(2a)+4 \\ \\ g(2a)=3(4a^2)-5(2a)+4 \\ \\ g(2a)=12a^2-10a+4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4uhqrjg47rlephzzncm1qehz593xo5nj2m.png)
ANSWERS:
• (a) g(-3) = 46
• (b) (-3, 46)
• (c) g(2a) = 12a² - 10a + 4