Question

How do you graph the parabola `y = x^2 - 6x + 7` using vertex, intercepts and additional points?

Answer 1

graph{x^2-6x+7 [-10, 10, -5, 5]}
Vertex: `(3,-2)`
x-intercepts: `(3+sqrt(2),0)`, `(3-sqrt(2),0)`
y-intercept: `(0,7)`

Explanation:

For a quadratic polynomial in standard form `f(x)=ax^2+bx+c` where `a!=0`:
1. Coordinate of the vertex: `(-b/(2a),c-b^2/(4a))`
2. y-intersect: `(0, c)`;
3. x-intersect: `((-b+-sqrt(b^2-4ac))/(2a),0)`. (There might be no x-intersect(s) for some quadratic polynomials, see this question for more detail. How do you solve quadratic equations algebraically? )

Evaluate these formulas with the exact value of polynomial to get the coordinate of the points.

Reference:https://en.wikipedia.org/wiki/Quadratic_function

Answer 2

Vertex: `(3,-2)`
x-Intercepts: `(1.59, 0)` and `(4.41, 0)`
y-Intercept: `(0, 7)`
Additional points: `(5,2)` and `(-5,62)`

Explanation:

The equation is in standard form
`y=ax^2+bx+c`

Where
`a=1`
`b=-6`
`c=7`

Because the coefficient of the `x^2` term `a=1` is positive (+1), the graph opens upward.

Only because `a=1`, the vertex determined by the following formula
`x_"vertex"=-b/(2a)=-(-6)/(2*1)=3`
`y_"vertex"=(3)^2-6(3)+7=-2`
Vertex: `(3,-2)`

The intercepts occur when `x=0` and `y=0`, respectively
When `x=0`, then `y=7`
When `y=0`, the quadratic equation gives us:

`x=(6+-sqrt((-6)^2-4(1)(7)))/(2(1))=(6+-sqrt(8))/(2)`
`x_1=(6-sqrt(8))/2~~1.59`
`x_2=(6+sqrt(8))/2~~4.41`

Using all these facts, you can sketch the graph. It should look like this:
graph{x^2-6x+7 [-1, 7, -3, 10]}