Question

Write a rule for a quadratic function that has x-intercepts of -3 and 4 with a y-intercept if 10. Show all work.

Answer 1

you know it looks like
\[f(x)=a(x+3)(x-4)\] and you know \[f(0)=10\] so you can solve for "a" via
\[a\times 3\times -4=10\]

Answer 2

x-intercept => y = 0,
x-intercept1 = -3 and x- intercept2 =4, means then that when y = 0, x = -3, we have (-3,0) or when y= 0, x = 4, (4,0)
y-intercept => x =0
(0,10)
y= ax^2 + bx + c =0, x= 4, x = -3 are roots, or solutions.
When x = 0, y= a * 0^2 + b * 0 +c, y = c. When x is 0, y is 10, then c = 10.

a(-3)^2 -3b + 10 = 0
9a -3 b = - 10

a (4)^2 + 4b + 10 = 0
16 a + 4 b = - 10

Answer 3

9a - 3b = - 10
16 a + 4b = - 10

9a - 3 b =16 a + 4b
- 7 a = 7 b
- a = b

Answer 4

ax^2 -ax + 10 = 0
â (x^2 - x) = - 10
a(16 - 4) = - 10
a * (12) = - 10
a = -5/6
b = 5/6

-5/6 x^2 + 5/6 x + 10 = 0

Answer 5

wow i think that is right but it is a lot of work.
you can write
\[f(x)=a(x+3)(x-4)\] and
\[10=f(0)=a\times 3\times -4\]
\[-12a=10\]
\[a=-\frac{5}{6}\]
so your answer is
\[f(x)=-\frac{5}{6}(x+3)(x-4)\] multiply out if you like