Question

The function f is defined by f(x)=x^2+3x-10 If f(x+5)=x^2+kx+30, k= Find the smallest zero of f(x+5), x=

Answer 1

f(x+5)=(x+5)^2+3(x+5)-10

f(x+5)=x^2+10x+25+3x+15-10

f(x+5)=x^2+13x+30 and f(x+5)=x^2+kx+30 so

k=13

Now factor x^2+13x+30

Find j and k such that jk=ac=30 and j+k+b=13 so j and k are 10 and 3 so

(x+3)(x+10)

So the two zeros occur when x=-3 and -10 the smallest of which is:

x=-10

Answer 2

f(x+5)
Step1:
f(x+5)=(x+5)^2+3(x+5)-10
f(x+5)=x^2+10x+25+3x+15-10
f(x+5)=x^2+13x+30

Step
f(x)=x^2+3x-10
f(x+5)=x^2+13x+30


f(x+5)=0

0=x^2+13x+30
factor
0=(x+3)(x+10)


x+3=0
x=-3

x+10=0
x=-10

-10<-3

smalle zero is x=-10.....