The function is in vertex form of a quadratic function.
Vertex form is f(x)=a(x-h)^2+k
The vertex is (h, k)
In this case, our vertex is (2, 2). The a value is positive which means that the graph points up. Since the graph points up, the vertex is the minimum, the lowest point of the graph. The range refers to y-values, so the range of the function is y/y>2. It can also be written as 2≤y<span>≤positive infinity.</span>
Its Letter A see photo for solution
Answer:
Step-by-step explanation:
A parallel line will have the same slope as the reference line. In this case, I don't see the "given line" as promised in the question. If it does appear, and it looks like y = 5x + 3, for example, the slope is 5 and the new line will have the same slope.
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<u>If this slope is correct</u>, we can start the equation for the parallel line that goes through point (-3,2) by starting with:</h3><h3 /><h3>y = 5x + b</h3><h3 /><h3>We need a value of b that forces the line to go through point (-3,2). We can do that by using the given point in the equation and solving for b:</h3><h3>y = 5x + b</h3><h3>2 = 5(-3) + b</h3><h3>b = 17</h3><h3 /><h3>The parallel line to y=5x+3 is</h3><h3>y = 5x + 17</h3><h3 /><h3>See attachment.</h3><h3 /><h3 /><h3 />
Answer:
Step-by-step explanation:
it is all of the side lengths of the four sides times how many trapezoids there are