Question

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 in the equation of the parabola

Answer 1

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

⋅

Explanation:

For complete solution see the above attachment!