The zeros of a parabola are -4 and 2 and (6,10) is a point on the graph which equation can be solved to determine the value of a
1 answer:
Answer:
Let consider the following linear equation systems by using the known points and second-grade polynomial:
(-1, 7)
a + b + c = 7a+b+c=7
(5, 7)
25\cdot a + 5\cdot b + c = 725⋅a+5⋅b+c=7
(6,10)
36\cdot a + 6 \cdot b + c = 1036⋅a+6⋅b+c=10
After some algebraic manipulation, the values for the polynomial coefficients are found:
a = \frac{3}{5}a= 53 , b = -\frac{18}{5}b=− 518
, c = 10c=10
The polynomial is:
y = \frac{3}{5}\cdot x^{2} - \frac{18}{5}\cdot x+10y= 53
⋅
Step-by-step explanation:
For complete solution see the above attachment!
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