Answer:
A, B, C
Step-by-step explanation:
Got it right on khan academy
A graph includes the ordered pair(2,4) write two different rules that would be possible for this graph. explain how you found yo
1 answer:
You might be interested in
Your calculator's cubic regression function can tell you the equation is
... f(x) = 2x³ + 5x² -12x = x(x +4)(2x-3)
The x-intercepts are -4, 0, +1.5.
_____
If you want to solve this "by hand", you can first of all recognize that since there is an x-intercept at 0, the cubic will only have three coefficients. That is, you can write the equation as
... f(x) = ax³ + bx² + cx
Substituting the given points (except (0, 0)) gives three linear equations in a, b, c.
... -a +b -c = 15 . . . . . for x=-1
... a + b + c = -5 . . . . for x=1
... 8a +4b +2c = 12 . . for x=2
adding the first two equations gives 2b=10, or b=5. Now, you can reduce the system to
... a + c = -10
... 4a +c = -4
Subtracting the first of these equations from the second gives 3a=6, or a=2. That tells you c=-12 (from a+c=10).
Then your equation is
... f(x) = x(2x² +5x -12)
Factoring by any of the usual techniques, or graphing, or using the quadratic formula will tell you the zeros (x-intercepts) are as above.
_____
Since the input values are sequential, you can also develop the function from differences of the output values. Those are 15, 0, -5, 12. First differences are -15, -5, +17. Second differences are +10, +22. The third difference is 12. Using the first of these differences in appropriate places in the interpolating polynomial formula, we have
... f(x) = 15 + (x+1)(-15 + (x)/2·(10 + (x-1)/3·(12))) = 2x³ +5x² -12x . . . . as above
The rest of the question is the attached figure.
============================================
Δ AYW a right triangle at Y ⇒⇒⇒ ∴ WA² = AY² + YW²
And AY = YB ⇒⇒⇒ ∴ WA² = YB² + YW² → (1)
Δ BYW a right triangle at Y ⇒⇒⇒ ∴ WB² = BY² + YW² → (2)
From (1) , (2) ⇒⇒⇒ ∴ WA = WB →→ (3)
Δ CXW a right triangle at Y ⇒⇒⇒ ∴ WC² = CX² + XW²
And CX = XB ⇒⇒⇒ ∴ WC² = XB² + XW² → (4)
Δ BXW a right triangle at Y ⇒⇒⇒ ∴ WB² = XB² + XW² → (5)
From (4) , (5) ⇒⇒⇒ ∴ WC = WB →→ (6)
From (3) , (6)
WA = WB = WC
given ⇒⇒⇒ WA = 5x – 8 and WC = 3x + 2
∴ <span> 5x – 8 = 3x + 2</span>
Solve for x ⇒⇒⇒ ∴ x = 5
∴ WB = WA = WC = 3*5 + 2 = 17
The correct answer is option D. WB = 17
============================================
Δ AYW a right triangle at Y ⇒⇒⇒ ∴ WA² = AY² + YW²
And AY = YB ⇒⇒⇒ ∴ WA² = YB² + YW² → (1)
Δ BYW a right triangle at Y ⇒⇒⇒ ∴ WB² = BY² + YW² → (2)
From (1) , (2) ⇒⇒⇒ ∴ WA = WB →→ (3)
Δ CXW a right triangle at Y ⇒⇒⇒ ∴ WC² = CX² + XW²
And CX = XB ⇒⇒⇒ ∴ WC² = XB² + XW² → (4)
Δ BXW a right triangle at Y ⇒⇒⇒ ∴ WB² = XB² + XW² → (5)
From (4) , (5) ⇒⇒⇒ ∴ WC = WB →→ (6)
From (3) , (6)
WA = WB = WC
given ⇒⇒⇒ WA = 5x – 8 and WC = 3x + 2
∴ <span> 5x – 8 = 3x + 2</span>
Solve for x ⇒⇒⇒ ∴ x = 5
∴ WB = WA = WC = 3*5 + 2 = 17
The correct answer is option D. WB = 17