Determine the number of pounds of nuts selling for $6 per pound and raisins selling for $3 per pound
1 answer:
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The quadratic function in vertex form is:
y = a(x - h)^2 + k
Where:
vertex = (h, k)
Axis of symmetry: x = h
The value of “a” determines whether the graph opens up or down, and makes the parent function wider or narrower.
The value of “h” determines how far left or right the parent function is translated.
The value of “k” determines how far up or down the parent function is translated.
Now that we have these definitions, we can substitute the given values into the vertex form to solve for “a”:
Use vertex = (-4, -1) and y-intercept, (0, 7):
7 = a(0+ 4)^2 - 1
7 = a(4)^2 - 1
7 = a(16) - 1
Add 1 to both sides:
7 + 1 = a(16) - 1 + 1
8 = 16a
Divide both sides by 16 to solve for “a”:
8/16 = 16a/16
1/2 = a
Since a = 1/2 (which is positive, implying that the parabola opens upward), and the vertex occurs at point (-4, -1) as the minimum point:
The quadratic equation in vertex form is:
y = 1/2(x + 4)^2 - 1
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y = a(x - h)^2 + k
Where:
vertex = (h, k)
Axis of symmetry: x = h
The value of “a” determines whether the graph opens up or down, and makes the parent function wider or narrower.
The value of “h” determines how far left or right the parent function is translated.
The value of “k” determines how far up or down the parent function is translated.
Now that we have these definitions, we can substitute the given values into the vertex form to solve for “a”:
Use vertex = (-4, -1) and y-intercept, (0, 7):
7 = a(0+ 4)^2 - 1
7 = a(4)^2 - 1
7 = a(16) - 1
Add 1 to both sides:
7 + 1 = a(16) - 1 + 1
8 = 16a
Divide both sides by 16 to solve for “a”:
8/16 = 16a/16
1/2 = a
Since a = 1/2 (which is positive, implying that the parabola opens upward), and the vertex occurs at point (-4, -1) as the minimum point:
The quadratic equation in vertex form is:
y = 1/2(x + 4)^2 - 1
Please mark my answers as the Brainliest, if you find my explanations/solution helpful :)
Answer: see below
<u>Step-by-step explanation:</u>
Any value x such that |x| < 1 will make the conjecture false
<em>In simpler words, let x be a fraction between 0 and 1.</em>
One example: Let x =
Then x⁴ =
=
<
so the conjecture is false.