User: desconectar
Maths Exercices

Vertex and Intercepts od a Quadratic Graph


The quadratic equation has an extreme value located at the vertex of the graph. This vertex point is either the highest point on the graph of the equation or the lowest point:


Finding the vertex of a parabola without graphing
The location of the coordinates of the vertex of the graph of y=ax2+bx+c depends only on the coefficients of a and b:

Coordinates of the vertex: (the y-value is found by substituting the x-value)

Find the vertex of f(x)=x2+6x+8
1. Find the x-coordinate of the vertex by using the formula , where a=1 and b=6:

2. Find the y-coordinate of the vertex by evaluating f(-3)


Hence, the vertex is at (-3,-1)


The intercepts of a parabola are where the curve or graph of the function crosses the axis.

Finding the intercepts of a parabola without graphing.
  • To solve for the x-intercepts of a quadratic function, you set f(x) equal to 0 (this is the same as setting y equal to 0)and then solve for x.
  • To solve for the y-intercept of a quadratic function, you set x equal to 0 and solve for f(x). A function has only one y-intercept.

Find the x-intercept(s) and the y-intercept(s) of the quadratic equation y=x2-4

Step 1. First, find the y-intercept. Since the y-intercept is the point where the graph crosses the y-axis, the value for x at this point is zero. Because we know that x=0, plug 0 in for x in the equation and solve for x.

Therefore, the y-intercept is (0,-4)

Step 2. Next, find the x-intercept(s). The x-intercept is the point, or points in this case, where the graph crosses the x-axis. In this case, we plug 0 in for y because y is always zero along the x-axis. Solve for x.


So x=-2 and x=2
The x-intercepts are (-2,0) and (2,0)

Step 3. To verify that the intercepts are correct, graph the equation on the coordinate plane:

Find the vertex and the intercepts of the quadratic equation: f(x)=-4x2+32x-60

( , ) ( , )
( , )
( , )