Answer:
see attached graph
Step-by-step explanation:
Given equation: [tex]y=x^2-4x+3[/tex]
Standard form of a quadratic equation: [tex]y=ax^2+bx+c[/tex]
where [tex]c[/tex] is the y-intercept
If [tex]a > 0[/tex] then the parabola opens upwards.
If [tex]a < 0[/tex] then the parabola opens downwards.
Therefore, the y-intercept of the graph is (0, 3)
To find the x-intercepts, factor the equation:
[tex]y=x^2-4x+3[/tex]
[tex]\implies y=x^2-x-3x+3[/tex]
[tex]\implies y=x(x-1)-3(x-1)[/tex]
[tex]\implies y=(x-3)(x-1)[/tex]
Therefore, the x-intercepts of the graph are (3, 0) and (1, 0)
To determine the vertex, differentiate the equation:
[tex]\implies \dfrac{dy}{dx}=2x-4[/tex]
Set to zero and solve for x:
[tex]2x-4=0 \implies x=2[/tex]
Substitute [tex]x=2[/tex] into the original equation and solve for y:
[tex]\implies y=(2)^2-4(2)+3=-1[/tex]
Therefore, the vertex (or turning point) is at (2, -1)
