Answer:
*Parabola that opens up
*y-intercept is 2
*x-intercepts are -1/3 and -2
*vertex is (-7/6 , -25/12)
Step-by-step explanation:
I will describe what this graph looks like.
First, the graph is quadratic because it is in the form ax² + bx + c. The function must be in the shape of a parabola. Parabolas look like a "U" shape.
Whether "a" is negative or positive represents of the parabola opens up or down. Since "3" is positive, the parabola opens up.
"c" represents the y-intercept. The y-intercept is when the graph touches the y-axis. The function must have a y-intercept of positive 2.
We can also find its x-intercepts, also called roots/zeroes, by substituting into the quadratic formula (Ignore the Â).
Using the form ax² + bx + c, we know:
a=3; b=7; c=2
Substitute into the formula.
Split the equation at the ± sign.
The graph has x-intercepts -2 and -1/3.
We can find the vertex of the graph. It is the part of the parabola that is the lowest (or highest, is it opens down).
Find the midpoint of the x-intercepts for the vertex x-coordinate:
(-1/3 - 2)/2 = (-1/3 - 6/3)/2 = (-7/3)*(1/2) = -7/6 = -1.167 = x
Substitute the vertex x-coordinate into the formula to find the "y" in vertex.
y=3x²+7x+2
y=3(-7/6)²+7(-7/6)+2
y= 147/36 + (-49/6) + 2
y= 147/36 + (-294/36) + 72/36
y= (147-294+72)/36
y = -75/36
y = -25/12 = -2.083
The vertex is (-7/6 , -25/12) OR about (-1.167, - 2.083).
The function represents a parabola. This is graphed by creating a series of (x,y) pairs and plotting them. The resultant shape is a parabola opening upwards.
The graph described by the function is a parabola as this is a quadratic function. To depict this graph, you plot specific points determined by varying the values of x and calculating the corresponding y values.
For instance, if x is 0, y would be 2 ( ), so (0,2) is one point on the graph. You continue this process with different x values to generate a series of (x,y) pairs, which you can then plot on a graph. Once you've plotted these points, you can connect them to see the graphical representation of the function.
In this case, you would get a parabola opening upwards since the coefficient of the term is positive.
#SPJ12
Answer;
= (1/8), (2/8), (3/8), (4/8), (5/8), (6/8), and (7/8)
The positive fractions less than 1, with an interval of 1/8 between each pair are;
1/8, 2/8, 3/8, 4/8, 5/8, 6/8, and 7/8
-To represent these fractions on the number line, subdivide the interval between0 and 1 into eight smaller intervals
B. 0/65
C. 11
D. 5/13
Answer:
c
Step-by-step explanation:
B. √ 64
2021 Edge :o)