Zero points of a function are its intersection points with the $X$-axis.

To find them, we set $Y=0$,

we get an equation that can sometimes be solved using a trinomial or the quadratic formula.

Zero points of a function are its intersection points with the $X$-axis.

To find them, we set $Y=0$,

we get an equation that can sometimes be solved using a trinomial or the quadratic formula.

**Two results -**

In this case, the function intersects the $X$-axis at two different points.**One result -**

In this case, the function intersects the $X$-axis at only one point, meaning the vertex of the parabola is exactly on the $X$-axis.**No results -**

In this case, the function does not intersect the $X$-axis at all, meaning it hovers above or below it.

The following function has been graphed below:

\( f(x)=-x^2+5x+6 \)

Calculate points A and B.

**Zero points describe a situation where the function equals zero.**

Let's look at an example:

We have the function

$X^2-4x+5$

We substitute into the quadratic formula and get:

${-4 \pm \sqrt{(-4)^2-4*1*5} \over 21}=-\frac{4\pm\sqrt{-4}}{2}$

This equation has no solution because the delta inside the root is negative. Therefore, this equation never equals zero, it hovers above the X-axis, and it has no zero points.

Test your knowledge

Question 1

The following function has been graphed below:

\( f(x)=x^2-6x+8 \)

Calculate points A and B.

Question 2

The following function has been graphed below:

\( f(x)=x^2-8x+16 \)

Calculate point A.

Question 3

The following function has been graphed below:

\( f(x)=x^2-6x+5 \)

Calculate points A and B.

\( \)

Determine the points of intersection of the function

$y=(x-5)(x+5)$

With the X

In order to find the point of the intersection with the X-axis, we first need to establish that Y=0.

0 = (x-5)(x+5)

When we have an equation of this type, we know that one of these parentheses must be equal to 0, so we begin by checking the possible options.

x-5 = 0

x = 5

x+5 = 0

x = -5

That is, we have two points of intersection with the x-axis, when we discover their x points, and the y point is already known to us (0, as we placed it):

(5,0)(-5,0)

This is the solution!

$(5,0),(-5,0)$

The following function has been graphed below:

$f(x)=-x^2+5x+6$

Calculate points A and B.

$(-1,0),(6,0)$

The following function has been graphed below:

$f(x)=x^2-6x+8$

Calculate points A and B.

$(2,0),(4,0)$

The following function has been graphed below:

$f(x)=x^2-8x+16$

Calculate point A.

$(0,16)$

The following function has been graphed below:

$f(x)=x^2-6x+5$

Calculate points A and B.

$(1,0),(5,0)$

Related Subjects

- Quadratice Equations and Systems of Quadraric Equations
- Quadratic Equations System - Algebraic and Graphical Solution
- Solution of a system of equations - one of them is linear and the other quadratic
- Intersection between two parabolas
- Word Problems
- Properties of the roots of quadratic equations - Vieta's formulas
- Ways to represent a quadratic function
- Various Forms of the Quadratic Function
- Standard Form of the Quadratic Function
- Vertex form of the quadratic equation
- Factored form of the quadratic function
- The quadratic function
- Quadratic Inequality
- Parabola
- Symmetry in a parabola
- Plotting the Quadratic Function Using Parameters a, b and c
- Methods for Solving a Quadratic Function
- Completing the square in a quadratic equation
- Squared Trinomial
- The quadratic equation
- Families of Parabolas
- The functions y=x²
- Family of Parabolas y=x²+c: Vertical Shift
- Family of Parabolas y=(x-p)²
- Family of Parabolas y=(x-p)²+k (combination of horizontal and vertical shifts)