Question

Find the value for a and write the equation if the x intercepts are (-3,0) and (5,0) and the point is (7.-16).

Answer 1

`\begin{gathered} y=(4)/(5)x^2-(8)/(5)x-12 \\ a=(4)/(5) \end{gathered}`

1) Given the intercepts and (5,0) and point (7,-16) let's plug it in

Let's set a system of linear equations to find a:

49a +7b +c

9a -3b +c

25a +5b +c

49a +7b +c=16

9a -3b +c=0 x (-1) To eliminate c

49a +7b +c =16

-9a +3b -c=0

49a +7b = 16 Adding both equations simultaneously

-9a +3b = 0

---------------------

40a +10b= 16

2) Now let's operate the other two equations, eliminating the c variable:

25a +5b +c =0

9a -3b +c= 16 x(-1)

25a +5b+c =0

-9a +3b -c=0

--------------------

16a +8b = 0

Now let's solve the system for a and b. Let's eliminate b:

40a +10b= 16 x 4

16a +8b = 0 x-5

160a +40b = 64

-80a -40b = 0

------------------------

80a = 64

a = 4/5

3) Now we can write this quadratic equation since the x-intercepts are x=-3 and x= 5

`\begin{gathered} y=a(x-x_1)(x-x_2) \\ y=(4)/(5)(x+3)(x-5) \\ y=(4)/(5)(x^2-5x\text{ +3x-15)} \\ y=(4)/(5)(x^2-2x-15) \\ y=(4)/(5)x^2-(8)/(5)x-12 \end{gathered}`