Answer:
see attached graph
Step-by-step explanation:
Given equation: ![y=x^2-4x+3](https://tex.z-dn.net/?f=y%3Dx%5E2-4x%2B3)
Standard form of a quadratic equation: ![y=ax^2+bx+c](https://tex.z-dn.net/?f=y%3Dax%5E2%2Bbx%2Bc)
where
is the y-intercept
If
then the parabola opens upwards.
If
then the parabola opens downwards.
Therefore, the y-intercept of the graph is (0, 3)
To find the x-intercepts, factor the equation:
![y=x^2-4x+3](https://tex.z-dn.net/?f=y%3Dx%5E2-4x%2B3)
![\implies y=x^2-x-3x+3](https://tex.z-dn.net/?f=%5Cimplies%20y%3Dx%5E2-x-3x%2B3)
![\implies y=x(x-1)-3(x-1)](https://tex.z-dn.net/?f=%5Cimplies%20y%3Dx%28x-1%29-3%28x-1%29)
![\implies y=(x-3)(x-1)](https://tex.z-dn.net/?f=%5Cimplies%20y%3D%28x-3%29%28x-1%29)
Therefore, the x-intercepts of the graph are (3, 0) and (1, 0)
To determine the vertex, differentiate the equation:
![\implies \dfrac{dy}{dx}=2x-4](https://tex.z-dn.net/?f=%5Cimplies%20%5Cdfrac%7Bdy%7D%7Bdx%7D%3D2x-4)
Set to zero and solve for x:
![2x-4=0 \implies x=2](https://tex.z-dn.net/?f=2x-4%3D0%20%5Cimplies%20x%3D2)
Substitute
into the original equation and solve for y:
![\implies y=(2)^2-4(2)+3=-1](https://tex.z-dn.net/?f=%5Cimplies%20y%3D%282%29%5E2-4%282%29%2B3%3D-1)
Therefore, the vertex (or turning point) is at (2, -1)