Question

How do you use the important points to sketch the graph of `y=x^2-7x+12`?

Answer 1

graph{x^2-7x+12 [-12.17, 19.86, -0.56, 15.45]}

Explanation:

Important Points of a Quadratic Function:
- vertex
- y-intercept
- x-intercept(s)

VERTEX
To find the vertex, complete the square and find `(h,k)`. See below.
`y=x^2-7x+12`
`y=(x^2-7x+49/4)+12-49/4`
`y=(x-7/2)^2-1/4 -> y=a(x-h)^2+k`
vertex: `(7/2,-1/4)`

X-INTERCEPT(S)
Factor the standard equation. Set each factor equal to zero. The values are the x-coordinates for the x-intercepts.
`y=x^2-7x+12`
`y=(x-4)(x-3)`
`x-4=0->x=4`
`x-3=0->x=3`
x-intercepts: (4,0), (3,0)

Y-INTERCEPT
Since the equation was given in standard form, we know that the c value in the equation is the y-coordinate of the y-intercept.
`y=x^2-7x+12->y=ax^2+bx+c`
`a=1`
`b=-7`
`c=12`
y-intercept: `(0,12)`

Now that you have all the important points, you can graph the parabola on an x and y plane.
graph{x^2-7x+12 [-12.17, 19.86, -0.56, 15.45]}