**Answer:**

Vertex: (3, -1). x-intercept(s): (4,0) & (2,0). y-intercept: (0,8). Domain: (-∞,∞). Range: [-1,∞) Axis of symmetry: x= 3

**Step-by-step explanation:**

the standard form of a parabola is ax² + bx + c which is also known as a **quadratic equation**

the vertex form of a parabola is a(x - h)² + k

**problem 1: x² - 6x + 8**

to solve this problem, we would need to put the equation into vertex form.

to find the vertex, we will use the formula <em>-b/2a, </em>where<em> b</em> is -6 and <em>a </em>is 1 to find <em>h (</em>also known as the x-coordinate). plug it in to the formula

-(-6)/2(1) = 6/2 = **3 <-- the x-coordinate **

to find the y-coordinate, we will plug in the newly found x-coordinate into the <em>original </em>equation

solve: x² - 6x + 8 when x = 3

(3)² - 6(3) + 8 = 9 - 18 + 8 = **-1 <-- the y-coordinate**

**so, the vertex of the equation is (3, -1)**

**next, we determine if the parabola opens up or down. **in this case, the parabola opens up because the coefficient infront of x² is **positive **(1). if the coefficient was **negative **(-1), the parabola would open down.

**now, we find the x & y intercepts. **you can plot these coordinates in your table of values.

**finding the x intercept(s): solve the equation for 0**

solving x² - 6x + 8 =0, we can factor this out to be (x - 4)(x - 2) = 0. using the zero-product rule, we solve each of those equations for 0, with x - 4 = 0 and x - 2 = 0. the results from these are 4 and 2.

this makes the x intercepts **(4, 0)** and** (2, 0)**

**finding the y intercept(s): substitute x = 0 in the original equation**

substituting 0 for x in the original equation gives us:

(0)² - 6(0) + 8, which gives us only 8 since the 0s cancel out everything else. you can also say that the **constant term **<em>**c**</em>** is the y intercept** in order to make things a bit easier.

this makes the y intercept **(0, 8)**

**finding the domain of a parabola:**

the domain of a parabola is <em>**always (-∞,∞)**</em>, because it continues from left to right indefinitely

**finding the range of a parabola:**

the range of a parabola goes from bottom to top. the **lowest y coordinate** that the parabola has is located at (3, -1), making the lowest y coodinate -1. since the parabola opens upward, we know that it continues to **positive infinity**.

the range of the parabola is [-1,∞). we use a bracket on -1 because the point is located on the graph and **is included in the solution**.

**the axis of symmetry: **the axis of symmetry is the x-coordinate of the vertex, which is 3. so, the axis of symmetry would be x=3.

**to find additional points**, plug in any real number into the original equation, and the output will be the y value. for example:

x² - 6x +8 when x = 1

(1)² - 6(1) + 8 -> 1 - 6 + 8= 3

so, another coordinate on the grid would be (1, 3).

with all of this information, you can graph the parabola starting at the vertex and going through the y & x intercept(s) and the oher points you may need on the graph, along with solving the other equation.

i have attached a roughly drawn parabola of what it should look similar to, listing the intercepts and the vertex.