Quadratics: 1F

What is a quadratic equation

An equation in which the highest exponent is 2 and $a$ cannot be zero(0).

The format of this equation looks like: $y = ax^2 + bx + c$

The shape of the parabola depends on the coefficient of x. If it is positive, then the graph will form a Upward Facing parabola. If the coefficient of x is negative, then the graph will form a Downward Facing parabola

Firstly, you need to find out intersection of points on x and y axis.
NOTE:
- The intersection of points with an x coordinate has a y-coordinate of 0 beacuse those points lie on the x-axis. In other words, if a point lies on the x-axis then it will have cordinates (x, 0)
- The intersection of points with an y coordinate has a x-coordinate of 0 beacuse those points lie on the y-axis. In other words, if a point lies on the y-axis then it will have cordinates (0, y)
Secondly, you need to find out the turing point of graph on y-axis
Lastly, you need to find out the y-coordinate of the turing point

Suppose you want plot a graph for $x^2 - 6x + 8$ .

Now, we would do is. We are going to find the intersection on y-axis.

 $0 = x(x - 4) - 2(x - 4)$

 $0 = (x - 4)(x - 2)$

 $x_1 = 4 \ \ | \ \ x_2 = 2$

Now we know the two intersecting points (4, 0) and (2, 0) for $y = 0$ . Now let's find intersection on x-axis.

 $y = 0^2 - 6(0) + 8$

Now let's find our turing point with the formual $\frac{-b}{2a}$ .

By using this equation we can calculate that turing point is $\frac{-(-6)}{2(1)}$ which is 3.

Now the last step is to find y-coordinate for the turning point. We will be substituting the value of turning point into our equation

 $y = 3^2 - 6(3) + 8$

Now we have everything we need. The graph will form a upward facing parabola.

Graph of x^2 - 6x + 8