Answer:
1.B) and 2. A)
Step-by-step explanation:
1.This question is very interesting......
Only by this statement, the question can be solved
f(x)=ax3−ax2+bx+4 is a multiple of x2+1;
Because the roots of the equation x2+1 would be the roots of f(x).
Therefore if j,k are roots of the equation then j+k=0 and j.k=1. Let the roots of f(x) be j,k,l.
Then,
j+k+l=−−aa
l=1
now, j.k.l=−4a
so, a=−4
now j.k+l(k+j)=b−4
Therefore b=−4
Finally a+b=−8
But f(4)=4(−43+42−4+1)
Therefore f(4)=−204
So the question may be incorrect.
2.The vertex of a parabola is at x = -b/2a
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your equation is:
-14x^2 + bx + c = y
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This equation is in standard form when y = 0.
We get:
-14x^2 + bx + c = 0
a = -14
b and c are unknown.
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Equation for the vertex is x = -b/2a
We know that we want the vertex at (6,0), so we are at the vertex when:
x = 6
y = 0
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substituting x = 6 in the equation for the vertex, we get:
6 = -b/2a
We know that a = -14, so this equation becomes:
6 = -b/(-28)
We solve for b to get:
b = 168
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Our equation now becomes:
-14x^2 + 168x + c = y
We know that the vertex is at (6,0) so we know that we want this equation to equal to 0 at the point x = 6.
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We set y = 0 and we substitute 6 for x in the equation to get:
-14*(6^2) + 168(6) + c = 0
This becomes:
-504 + 1008 + c = 0
Thie becomes:
504 + c = 0
we solve for c to get:
c = -504
Our equation becomes:
y = -14x^2 + 168x - 504
Answer to your question is:
b = 168
c = -504
Graph of this equation is shown below:
Because a is negative, this graph opens downward.
If a was positive, it would open upward.
