7/5y= y+2/5y
0.68y = y-0.32y
3/5y=y-2/5y
1.32y=y+0.32y
For simplicity, I’m marking the six different terms on the main part of the image as 1-6 and the six equations/points on the bottom of the image as A-F.
First, we can look at the bottom and see that there are four equations, but A and F are both points. Looking at the terms above, the only ones that are points are 1 and 6. Since the y-intercept is when a function passes through the y-axis, that means that the x value of the point must be 0. Therefore, A matches with 6.
That leaves F to match with 1. The equation (-b/2a) is what is used in a quadratic equation to locate the vertex, so it makes sense that the vertex would be when you put that number into the equation.
Now on to the four different equations. One of the equations, C, is calculating an x value. This means that it has to match with 2 because it’s drawing a line vertically through the point we already determined to be the vertex of a parabola in the above paragraph.
For the three functions that remain, B has no numbers in it, only the standard letters as placeholders for various numbers you would put into a quadratic function. Therefore, it has to match with 3 since it’s just the standard equation.
That leaves us with D and E to match with 4 and 5, meaning that one of them has to be a negative equation and the other has to be a positive equation. This is where you look at the a value in the function. If the a value is positive, then the quadratic opens up (with the tip pointed down) while if the a value is negative, then the quadratic opens down (with the tip pointed up). This means that D matches with 4 and E matches with 5.
In conclusion,
1 = F
2 = C
3 = B
4 = D
5 = E
6 = A
Hope that helped!
Answer:
Step-by-step explanation:
hello : y= 4x² + 5x + 2
the x-intercepts when y=0 means : 4x² + 5x + 2=0
a=4 b= 5 c=2
∆ = b²- 4ac ∆ = (5)²- 4(4)(2) =25-32 = -7
∆ < 0 no reals solutions no x-intercepts
Answer:
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