Answer:
To draw this graph, we start from the left in quadrant 3 drawing the curve to -4 on the x-axis to touch it but not cross. We continue back down and curve back around to cross the x-axis at -1. We continue up past -1 and curve back down to 5 on the x-axis. We touch here without crossing and draw the rest of our function heading back up. It should form a sideways s shape.
Step-by-step explanation:
A polynomials is an equation with many terms whose leading term is the highest exponent known as degree. The degree or exponent tells how many roots exist. These roots are the x-intercepts.
This polynomial has roots -4, -1, and 5. This means the graph must touch or cross through the x-axis at these x-values. What determines if it crosses the x-axis or the simple touch it and bounce back? The even or odd multiplicity - how many times the root occurs.
In this polynomial:
Root -4 has even multiplicity of 4 so it only touches and does not cross through.
Root -1 has odd multiplicity of 3 so crosses through.
Root 5 has even multiplicity of 6 so it only touches and does not cross through.
Lastly, what determines the facing of the graph (up or down) is the leading coefficient. If positive, the graph ends point up. If negative, the graph ends point down. All even degree graphs will have this shape.
To draw this graph, we start from the left in quadrant 3 drawing the curve to -4 on the x-axis to touch it but not cross. We continue back down and curve back around to cross the x-axis at -1. We continue up past -1 and curve back down to 5 on the x-axis. We touch here without crossing and draw the rest of our function heading back up. It should form a sideways s shape.
There isn't anything there
Answer:
The answer is "NOT"
Step-by-step explanation:
- To demonstrate a similar statement we want to have yet another side or some other angle, however, the third section is not proportional.
- Its sides were equivalent in proportions (yeah sure they are) + corner by the by is not given.
- There is no triangle angle provided so, that we can not enforce the SAS argument.
- It is not evidence of similarity between ABC ≈DEF.