Question

Sketch the graph of y=x^2-4x+3

Answer 1

see attached graph

Explanation:

Given equation:
`y=x^2-4x+3`

Standard form of a quadratic equation:
`y=ax^2+bx+c`

where
`c` is the y-intercept

If
`a > 0` then the parabola opens upwards.

If
`a < 0` 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`

`\implies y=x^2-x-3x+3`

`\implies y=x(x-1)-3(x-1)`

`\implies y=(x-3)(x-1)`

Therefore, the x-intercepts of the graph are (3, 0) and (1, 0)

To determine the vertex, differentiate the equation:

`\implies (dy)/(dx)=2x-4`

Set to zero and solve for x:

`2x-4=0 \implies x=2`

Substitute
`x=2` into the original equation and solve for y:

`\implies y=(2)^2-4(2)+3=-1`

Therefore, the vertex (or turning point) is at (2, -1)