We will graph this equation by finding and plotting its intercepts, then connecting them with a smooth curve. To find the and , we will need to substitute 0 for one variable, solve, then repeat for the other variable.
Finding the x-intercepts
To find where a function , the function can be set equal to zero. Then, the x-values that satisfy the equation are the zeros of the function, also called the roots. Thus, to find the x-intercepts of the given function we can set it equal to zero and solve for x.
2x^2+6x+4=0
To solve the above equation for x, we can start by factoring. Then we will use the .
Factoring
Let's start factoring by first identifying the .
Factor Out The GCF
The GCF of an expression is a common factor of the terms in the expression. It is the common factor with the greatest coefficient and the greatest . In this case, the GCF is 2.
2x^2+6x+4=0
2* x^2+ 2* 3x+ 2* 2=0
2(x^2+3x+2)=0
The result of factoring out a GCF from the given equation is a with a leading coefficient of 1.
2( x^2+3x+2)=0
Let's temporarily only focus on this trinomial, and we will bring back the GCF after factoring.
Factor the Expression
To factor a trinomial with a leading coefficient of 1, think of the process as multiplying two in reverse. Let's start by taking a look at the constant term.
x^2+3x+2=0
In this case we have 2. This is a positive number, so for the product of the constant terms in the factors to be positive these constants must have the same sign (both positive or both negative.)
| Factor Constants
|
Product of Constants
|
| 1 and 2
|
2
|
| -1 and -2
|
2
|
Next, let's consider the coefficient of the linear term.
x^2+3x+2=0
For this term, we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, 3.
| Factors
|
Sum of Factors
|
| 1 and 2
|
3
|
| -1 and -2
|
-3
|
We found the factors whose product is 2 and whose sum is 3.
x^2+3x+2=0
⇕
(x+1)(x+2)=0
Wait! Before we finish, remember that we factored out a GCF from the original equation. To fully complete the factored equation, let's reintroduce that GCF now.
2(x+1)(x+2)=0
Now the equation is written in a factored form.
Zero Product Property
Since the equation is already written in factored form, we can now use the Zero Product Property.
2(x+1)(x+2)=0
(x+1)(x+2)=0
lcx+1=0 & (I) x+2=0 & (II)
lx=- 1 x+2=0
lx_1=- 1 x_2=- 2
We found that x=- 1 or x=- 2.
Finding the y-intercept
To find the y-intercept, consider the point where the graph of the equation crosses the y-axis. The x-value of the ( x, y) coordinate pair at the y-intercept is 0. Therefore, substituting 0 for x will give us the y-intercept.
y=2x^2+6x+4
y=2( 0)^2+6( 0)+4
y=2(0)+6(0)+4
y=4
A y-intercept of 4 means that the graph passes through the y-axis at the point (0, 4).
Finding the Vertex
We have already found all the intercepts of the given parabola. To graph it properly, we should also find the vertex.
The x-coordinate of the vertex is exactly halfway between x-intercepts. Since we know that x_1=- 1 and x_2=- 2, the x-coordinate of the vertex is halfway between (- 1,0) and (- 2,0).
x=x_1+x_2/2 ⇒ x=- 1+( - 2)/2=- 3/2
We can see that the x-coordinate of the vertex is - 32. To find the y-coordinate, we will substitute - 32 for x in the given equation.
y=2x^2+6x+4
y=2( - 3/2)^2+6( - 3/2)+4
y=2(9/4)+6(- 3/2)+4
y=2(9/4)-6(3/2)+4
y=18/4-18/2+4
y=9/2-18/2+4
y=- 9/2+4
y=- 9/2+8/2
y=- 1/2
The y-coordinate of the vertex is - 12. Therefore, the vertex is the point (- 32,- 12).
Graphing the Equation
We can now graph the equation by plotting the intercepts and vertex and connecting them with a line.
