Answer:
The y-intercept is: (0, 15)
The graph is also attached below.
Explanation:
Given
![f\left(x\right)=-3\left(x-2\right)^2+27](https://img.qammunity.org/2021/formulas/mathematics/middle-school/oso0pm3q67fpzntz67s1zpeb6sopm21wgu.png)
Determining the y-intercept
As we know that y-intercept is the point on a graph where x = 0
so
Putting x = 0 in the function
![y=-3\left(0-2\right)^2+27](https://img.qammunity.org/2021/formulas/mathematics/middle-school/ie7pth1ejtclx05xbk3ewzvpjk2vlxfacw.png)
![y=-2^2\cdot \:3+27](https://img.qammunity.org/2021/formulas/mathematics/middle-school/b26b81n8ifgali5rshseyd5zuvwtxr0nk3.png)
![y=-12+27](https://img.qammunity.org/2021/formulas/mathematics/middle-school/bvva2kiq2e2r2u9hdhjx5mgdbu4o8o1s7p.png)
![\mathrm{Add/Subtract\:the\:numbers:}\:-12+27=15](https://img.qammunity.org/2021/formulas/mathematics/middle-school/osa6j2bbw7m63chz0uhwgoao58y8ip40pb.png)
![y=15](https://img.qammunity.org/2021/formulas/mathematics/high-school/66rzi8jzu2mc7dxjywvuvjahhpxa1eoagc.png)
Therefore, the y-intercept is: (0, 15)
The graph is also attached below.