Answer:
She made the mistake in step 2 when she applied the commutative property. That was unnecessary
Y2-y1/x2-x1
5-(-3)/(-6-(-1))
8/-5
-8/5 is the slope
Answer:
(-1, 2) — derivative is zero
Step-by-step explanation:
A critical point of a function is a point at which the slope is <em>zero</em> or <em>undefined</em>. (The critical point must actually be in the domain of the function. Vertical asymptotes are <em>not</em> critical points.)
You find the answer by looking for points on the graph where a tangent to the function is horizontal or vertical.
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In the given graph, the curve approaches horizontal near the point (-1, 2), so it is reasonable to estimate that that point is a critical point.
The attached graph also shows the inverse function. That has a critical point at (2, -1), where the slope is undefined (a tangent to the graph is vertical).
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The graph of a circle or ellipse does not pass the vertical line test, so is not a function. However, the graph still has critical points where the tangents are vertical or horizontal. The second attachment shows those critical points.
<span>Graph the first equation (number of baskets made. You can do this by hand by picking a value for x (number of 2-point baskets) and solving for y. Let’s pick 2 for x.
x = 2, y =
Therefore, the point (2, ) is on the graph.
Let’s pick another value for x to find a second point on the graph. Let x be 4.
x = 4, y =
Therefore, the point (4, ) is on the graph.
Let’s pick one more value for x. Let x be 6.
x = 6, y =
Therefore, the point (6, ) is on the graph. Graph these three points and sketch the line through them.</span>
Answer:
ok
Step-by-step explanation: