Zero points of a function are its intersection points with the -axis.
To find them, we set ,
we get an equation that can sometimes be solved using a trinomial or the quadratic formula.
Finding the Zeros of a Parabola
Finding the zeros of a quadratic function of the form \(y=ax^2+bx+c\)
When trying to find the zero point, you can encounter three possible results:
- Two results -
In this case, the function intersects the -axis at two different points. - One result -
In this case, the function intersects the -axis at only one point, meaning the vertex of the parabola is exactly on the -axis. - No results -
In this case, the function does not intersect the -axis at all, meaning it hovers above or below it.
Test yourself on zeros of a fuction!
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
We substitute into the quadratic formula and get:
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.
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Test your knowledge
Question 1
The following function has been graphed below:
\( f(x)=x^2-6x+5 \)
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:
\( f(x)=x^2-8x+16 \)
Calculate point C.
Examples with solutions for Zeros of a Fuction
Exercise #1
The following function has been graphed below:
Calculate point C.
Video Solution
Step-by-Step Solution
To solve the exercise, first note that point C lies on the X-axis.
Therefore, to find it, we need to understand what is the X value when Y equals 0.
Let's set the equation equal to 0:
0=x²-8x+16
We'll use the preferred method (trinomial or quadratic formula) to find the X values, and we'll discover that
X=4
Answer
Exercise #2
Determine the points of intersection of the function
With the X
Video Solution
Step-by-Step Solution
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!
Answer
Exercise #3
The following function has been graphed below:
Calculate points A and B.
Video Solution
Answer
Exercise #4
The following function has been graphed below:
Calculate points A and B.
Video Solution
Answer
Exercise #5
The following function has been graphed below:
Calculate point A.
Video Solution
Answer
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)
