In algebra when we talk about symmetry, we are talking about a line that divides a figure exactly in half, so when we are working with quadratic functions, we had already mentioned that its graph will always be a parabola, therefore the axis of symmetry will be the line that divides the parabola in half, it's as if that axis were a mirror and the part on the right side reflected on the left side and vice versa.
If you are interested, you can go to the following link for more information on the topic: Symmetry
The functions y=x²
In the general form of a quadratic function, we mention that there are three terms: the quadratic term, linear term, and the constant term, to which we can say it is a complete quadratic function for having all its terms, however, this will not always be the case, that is we can have an incomplete quadratic function where it may lack the linear term or the constant term or even both terms, an incomplete function of the form y=x2 means that b=0 and c=0. This is the most basic form of a quadratic function and its graph will be a parabola with the vertex at the origin of the Cartesian plane, that is, at (0,0).
If you are interested, you can go to the following link for more information on the topic: The functions y=x²
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Test your knowledge
Question 1
\( y=-x^2+x+5 \)
Incorrect
Correct Answer:
\( a=-1,b=1,c=5 \)
Question 2
\( y=3x^2+3x-4 \)
Incorrect
Correct Answer:
\( a=3,b=3,c=-4 \)
Question 3
\( y=-4x^2+3 \)
Incorrect
Correct Answer:
\( a=-4,b=0,c=3 \)
Family of Parabolas y=x²+c: Vertical Shift
In the case of the family of parabolas y=x2+C, we now have the quadratic term and the constant term, graphically the constant term, that is C, will indicate whether the graph of the parabola shifts up or down on the Y axis depending on the value of C.
Unlike the previous parabolas, now the term p will indicate that the parabola will move horizontally, only if p>0 then the parabola will move to the left (negative X axis) the value of p; if p<0, that is, negative, it will now move to the right (positive X axis).
If you are interested, you can go to the following link for more information on the topic: Family of parabolas y=(x-p)²
Do you know what the answer is?
Question 1
\( y=-3x^2-4 \)
Incorrect
Correct Answer:
\( a=-3,b=0,c=-4 \)
Question 2
\( y=-5x^2 \)
Incorrect
Correct Answer:
\( a=-5,b=0,c=0 \)
Question 3
\( y=-x^2+3x+40 \)
Incorrect
Correct Answer:
\( a=-1,b=3,c=40 \)
Family of parabolas y=(x-p)²+k (combination of horizontal and vertical shifts)
This family of parabolas is the combination of the two previous ones, that is; The parabola will also move up or down (vertically) depending on the value of K. And p will indicate if it moves to the right or left (horizontally).
As we well know, factoring is writing a term as a multiplication, in the case of a quadratic function it is similar, the factored form will be written as a multiplication and in this case it helps us find the intersections of the parabola on the X axis. In other words, we will find the points where the parabola intersects with the X axis. The factored form can be observed as follows:
Finding the zeros of a quadratic function through its form
A quadratic function can be written in the following way:
f(x)=ax2+bx+c=0
We can observe that the function is equal to zero, therefore we can find the zeros, graphically they will be the intersections of the parabola with the X axis. There are various ways to find these zeros, one of them will be by the factoring method or by the quadratic formula method.
Completing the square in a quadratic equation
We already mentioned that a quadratic function can be in complete or incomplete form, the latter being those when it lacks one of the terms, either the linear term or the independent term, but never the quadratic term, since without this term it is no longer a quadratic function.
When we have an incomplete equation, in the case where we only have the quadratic term and linear term, we can complete the square through a series of steps. This is, find a number in such a way that we have all three terms and that it can be factored by the perfect square trinomial and thus find the solutions to the equation.
To find the solution of a quadratic equation, we can find it by factorization or by the quadratic formula, also called the general formula. This formula will help us find the solutions or roots that satisfy the equality in the equation. Graphically, we will find where it cuts the X axis or the intersections of the parabola with said axis. The coefficients of the terms that make up the quadratic function tell us certain things in the graph of the parabola.
From the quadratic equation ax2+bx+c=0
a, b, and c are coefficients or in this case, we will call them parameters. The quadratic formula that will help us find the roots is:
X1,2=2a−b±b2−4ac
A quadratic equation can have at most two solutions.
If you are interested, you can go to the following link for more information on the topic: The quadratic formula
Do you know what the answer is?
Question 1
\( y=-x^2+x+5 \)
Incorrect
Correct Answer:
\( a=-1,b=1,c=5 \)
Question 2
\( y=3x^2+3x-4 \)
Incorrect
Correct Answer:
\( a=3,b=3,c=-4 \)
Question 3
\( y=-4x^2+3 \)
Incorrect
Correct Answer:
\( a=-4,b=0,c=3 \)
Equations and systems of equations of second degree or quadratic
Just as in the case of a linear equation we encounter systems of linear equations where we study various methods to find the solution to the system, in the case of quadratic equations we can also work with systems of quadratic equations where we must also find the values of X and Y in such a way that the solutions satisfy both equations. Graphically, this system of equations can be observed as two parabolas where we find the coordinates of the two intersection points in the case of having two solutions, one intersection point when having a single solution, and lastly no intersection point when the system has no solution.
Quadratic Equations System - Algebraic and Graphical Solution
In a system of quadratic equations, we can find its solution algebraically and also graphically as we did with systems of linear equations. When we find the solution to a system of quadratic equations by the algebraic method, we can do it by equating, while graphically we can observe it with the points of intersection between the two parabolas, and we say that two systems of equations:
Have a unique solution when the parabolas only intersect at a single point.
Have two solutions when it can be observed that there are two points of intersection
And there is no solution when there is no point of intersection.
Solution of a system of equations when one of them is linear and the other quadratic
We may encounter systems of two equations where one of them is linear and the other is quadratic. We will solve this type of systems by the substitution method, where we will solve for the variable y in the linear equation and substitute this value into the quadratic, and then solve the resulting quadratic equation using the general formula. The solution or solutions will be the intersections between a straight line and a parabola.
As we previously mentioned, we can encounter three types of cases when we want to find the solution to a system of two quadratic equations. When these two equations are graphed on the Cartesian plane, it will guide us to find the solution:
There is a unique solution when the parabolas intersect at only one point.
There are two solutions when it can be observed that there are two points of intersection
And there is no solution when there is no point of intersection.
Do you think you will be able to solve it?
Question 1
\( y=3x^2-81 \)
Incorrect
Correct Answer:
\( a=3,b=0,c=-81 \)
Question 2
\( y=x^2 \)
Incorrect
Correct Answer:
\( a=1,b=0,c=0 \)
Question 3
\( y=2x^2+3 \)
Incorrect
Correct Answer:
\( a=2,b=0,c=3 \)\( \)
Word Problems
When we have problem applications that involve quadratic equations, we must use algebraic language, What does this mean? When we are presented with these types of problems, we must formulate our equations using the statements provided to us in such a way that we must translate this normal language into algebraic language and structure the equations, in this case, two equations to form our system of equations and subsequently find the solution by any method studied so far, and once the solution to the system of equations is obtained, now find the correct and viable solution to the initial problem.
If you are interested, you can go to the following link for more information on the topic: Verbal problems
Quadratic Inequality
Now we will talk about an inequality, that is, something that is not equal, and in this case, we will not have one or two solutions, but rather a solution set, where the solution will indicate in which range of solutions will make a quadratic equation positive or negative. In the following way:
ax2+bx+c<0
ax2+bx+c>0
If you are interested, you can go to the following link for more information on the topic: Quadratic inequality
Test your knowledge
Question 1
\( y=x^2+10x \)
Incorrect
Correct Answer:
\( a=1,b=10,c=0 \)
Question 2
\( y=x^2-6x+4 \)
Incorrect
Correct Answer:
\( a=1,b=-6,c=4 \)
Question 3
\( y=2x^2-5x+6 \)
Incorrect
Correct Answer:
\( a=2,b=-5,c=6 \)
Ways to Present a Quadratic Function
A quadratic function can be presented in the following ways:
Algebraically: f(x)=ax2+bx+c
Numerically: tabulation
Graphically: Parabola.
If you are interested, you can go to the following link for more information on the topic: Ways to present a quadratic function
Ways to Represent a Function
A function can be presented in four ways:
Verbally
Algebraically (Relation function)
Numerically (tabulation)
Graphically
Examples and exercises with solutions of the quadratic function
Exercise #1
y=x2+x+5
Video Solution
Step-by-Step Solution
To solve this problem, we will identify the parameters of the given quadratic function step-by-step:
Step 1: Define the problem statement: We have y=x2+x+5.
Step 2: Identify the standard form of a quadratic function, which is y=ax2+bx+c.
Step 3: Compare the given quadratic expression with the standard form to identify the coefficients.
Now, let's analyze the quadratic function provided:
From the given expression y=x2+x+5:
- The coefficient of x2 is 1, so a=1.
- The coefficient of x is 1, so b=1.
- The constant term is 5, so c=5.
Therefore, the parameters of the quadratic function are a=1, b=1, and c=5.
Consequently, the correct choice from the provided options is (a=1,b=1,c=5).
Answer
a=1,b=1,c=5
Exercise #2
y=−x2+x+5
Video Solution
Step-by-Step Solution
To solve the problem of identifying the coefficients in the quadratic function y=−x2+x+5, we follow these steps:
Step 1: Write down the general form of a quadratic equation: y=ax2+bx+c.
Step 2: Compare the given equation y=−x2+x+5 to the general form.
Step 3: Identify the value of each coefficient:
The coefficient of x2 is −1, so a=−1.
The coefficient of x is +1, so b=1.
The constant term is +5, so c=5.
Therefore, the parameters of the quadratic function are a=−1, b=1, and c=5.
This matches choice 2, confirming the parameters in the quadratic function.
Final Answer:a=−1,b=1,c=5.
Answer
a=−1,b=1,c=5
Exercise #3
y=3x2+3x−4
Video Solution
Step-by-Step Solution
To solve this problem, we'll follow these steps:
Step 1: Compare the given quadratic function with the standard form.
Step 2: Directly identify the coefficients a, b, and c.
Step 3: Verify the correct choice from the provided options, if applicable.
Now, let's work through each step:
Step 1: The given quadratic function is y=3x2+3x−4. The standard form for a quadratic equation is y=ax2+bx+c.
Step 2: By comparing the given equation to the standard form, we can identify the coefficients:
- a=3, from the term 3x2.
- b=3, from the term 3x.
- c=−4, from the constant term −4.
Step 3: With these values, compare them to the given choices. The choice that matches these values is option 3: a=3,b=3,c=−4.
Therefore, the solution to the problem is a=3,b=3,c=−4.
Answer
a=3,b=3,c=−4
Exercise #4
y=−4x2+3
Video Solution
Step-by-Step Solution
To solve this problem, we'll compare the given quadratic function with its standard form:
Step 1: Recognize the given function as y=−4x2+3.
Step 2: Write down the standard form of a quadratic function, which is y=ax2+bx+c.
Step 3: Match corresponding terms to identify a, b, and c.
Now, let's work through these steps:
Step 1: The given function is y=−4x2+3.
Step 2: The standard form of a quadratic function is y=ax2+bx+c.
Step 3: By direct comparison:
- The coefficient of x2 in the given expression is −4. Therefore, a=−4.
- There is no x term in the given expression, which implies the coefficient b=0.
- The constant term in the given expression is 3, indicating c=3.
Therefore, the solution is a=−4, b=0, c=3, which matches with choice 3.
Answer
a=−4,b=0,c=3
Exercise #5
y=−3x2−4
Video Solution
Step-by-Step Solution
To solve this problem, we will follow these steps:
Step 1: Identify the form of a standard quadratic equation.
Step 2: Compare the given function with the quadratic standard form.
Step 3: Match the coefficients to the given answer choices.
Now, let's work through each step:
Step 1: The standard form of a quadratic function is y=ax2+bx+c.
Step 2: Given the function y=−3x2−4, we compare this with the standard form:
Coefficient a is associated with x2. Here, a=−3.
Coefficient b is associated with x. Since there is no x term, b=0.
The constant term c is the standalone number, which is c=−4.
Step 3: Given the coefficients a=−3, b=0, and c=−4, match these with the choices provided. The correct choice is Choice 4: a=−3,b=0,c=−4.
Therefore, the solution to the problem is a=−3,b=0,c=−4.