Question

How do you use the important points to sketch the graph of `F(x)=x²-4x+9`?

Answer 1

Refer to the explanation.

Explanation:

Given:

`f(x)=x^2-4x+9` is a quadratic equation in standard form:

`f(x)=ax^2+bx+c`,

where:

`a=1`, `b=-4`, `c=9`

Determine the discriminant `(b^2-4ac)` and its relationship to zero.

`b^2-4ac=`

`(-4^2-4*1*9)=`

`-52`

Since the discriminant is less than zero, the parabola does not touch the x-axis, so there are no roots. We will need to determine the vertex, y-intercept, and additional points.

Vertex: maximum or minimum point on the parabola

The x-coordinate is determined using the formula `x=(-b)/(2a)`.

`x=(-(-4))/(2*1)`

`x=4/2`

`x=2`

To find the y-coordinate of the vertex, substitute `y` for `f(x)` and substitute `2` for `x`.

`y=2^2-4(2)+9`

`y=4-8+9`

`y=5`

The vertex is `(2,5)`.

Y-intercept: value of `y` when `x=0`

Substitute `0` for `x` in the equation and solve for `y`.

`y=0^2-4(0)+9`

`y=9`

The y-intercept is `(0,9)`.

Additional points: select values for `x` and solve for `y`.

Point 1: `x=4`

`y=4^2-4(4)+9`

`y=16-16+9`

`y=9`

Point 1 is `(4,9)`

Point 2: `x=-1`

`y=-1^2-4(-1)+9`

`y=1+4+9`

`y=14`

Point 2 is `(-1,14)`

Point 3: `x=5`

`y=5^2-4(5)+9`

`y=25-20+9`

`y=14`

Point 3 is `(5,14)`

Plot the vertex, y-intercept, and the three additional points. Sketch a parabola through the points. Do not connect the dots.

graph{y=x^2-4x+9 [-10.3, 9.7, 5.72, 15.72]}