The smallest positive integer that the intermediate value theorem guarantees a zero exists between 0 and a is 3.
What is the intermediate value theorem?
Intermediate value theorem is theorem about all possible y-value in between two known y-value.
x-intercepts
-x^2 + x + 2 = 0
x^2 - x - 2 = 0
(x + 1)(x - 2) = 0
x = -1, x = 2
y intercepts
f(0) = -x^2 + x + 2
f(0) = -0^2 + 0 + 2
f(0) = 2
(Graph attached)
From the graph we know the smallest positive integer value that the intermediate value theorem guarantees a zero exists between 0 and a is 3
For proof, the zero exists when x = 2 and f(3) = -4 < 0 and f(0) = 2 > 0.
<em>Your question is not complete, but most probably your full questions was</em>
<em>Given the polynomial f(x)=− x 2 +x+2 , what is the smallest positive integer a such that the Intermediate Value Theorem guarantees a zero exists between 0 and a ?</em>
Thus, the smallest positive integer that the intermediate value theorem guarantees a zero exists between 0 and a is 3.
Learn more about intermediate value theorem here:
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Answer:
B. (-0.73, 0), (2.73, 0)
Step-by-step explanation:
vertex form of a parabola
y = a(x-h)^2 +k
y = a(x-1)^2 -9
substitute the point in to find a
-6 = a (0-1) ^2 -9
-6 = a *1 -9
add 9 to each side
-6+9 = a -9+9
3 =a
y = 3(x-1)^2 -9
FOIL
y = 3(x-1)(x-1) -9
y = 3(x^2-2x+1) -9
distribute
y = 3x^2-6x+3-9
combine like terms
y = 3x^2 -6x -6
factor out 3
y= 3(x^2 - 2x -2)
set = 0
0 = 3(x^2 - 2x -2)
x^2 - 2x -2 =0
using the quadratic formula
-b±sqrt(b^2 -4ac)
----------------------
2a
-(-2) ± sqrt(2^2 - 4(1)(-2))
-------------------------------
2(1)
2 ± sqrt(4+8)
------------------
2
2±sqrt(12)
---------------
2
2±2sqrt(3)
----------------
2
1±sqrt(3)
roots: 2.73, and -.73
Answer:
x=15
Step-by-step explanation:
15+5=20
20/2=10
So, 15 =x
I can’t see A-c can u take another picture
Point slope: y-y1 = m(x-x1)
Plug in the numbers
Solution: y+2 = -4(x + 9)