Question

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

Answer 1

Answer:

see attached graph

Step-by-step 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:

$\implies \dfrac{dy}{dx}=2x-4$

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)