SOLUTION: Find the value of a and b so that the graph of y=sqrt(ax + b) passes through points (-4,6) and (1,4).

Algebra ->  Graphs -> SOLUTION: Find the value of a and b so that the graph of y=sqrt(ax + b) passes through points (-4,6) and (1,4).      Log On


   

Question 1091721: Find the value of a and b so that the graph of y=sqrt(ax + b) passes through points (-4,6) and (1,4).
Found 2 solutions by Fombitz, MathLover1:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
6=sqrt%28a%28-4%29%2Bb%29
6=sqrt%28b-4a%29
b-4a=36
and
4=sqrt%28a%281%29%2Bb%29
4=sqrt%28a%2Bb%29
a%2Bb=16
So then subtracting,
a%2Bb-%28b-4a%29=16-36
a%2Bb-b%2B4a=-20
5a=-20
Solve for a then use either equation to solve for b.

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

In triangle ABC, median AM is such that angle BAC is divided in ratio 1:2. AM is extended through M to D so that angle DBA is a right angle, then the ratio AC : AD is equal to

a median is a line joining a vertex with the mid-point of the opposite side
to find the value of a+ and b so that the graph of y=sqrt%28ax+%2B+b%29 use given points (-4,6) and (1,4)

y=sqrt%28ax+%2B+b%29 ...for (-4,6)
6=sqrt%28a%28-4%29+%2B+b%29
6=sqrt%28-4a+%2B+b%29 ....square both sides
36=-4a+%2B+b...solve for b
36%2B4a+=+b....eq.1

y=sqrt%28ax+%2B+b%29 ...for (1,4)
4=sqrt%28a%281%29+%2B+b%29
4=sqrt%28a+%2B+b%29 ....square both sides
16=a+%2B+b...solve for b
16-a+=+b....eq.2
from eq.1 and eq.2 we have
36%2B4a+=+16-a+..solve for a
a%2B4a+=+16-36+
5a+=+-20+
highlight%28a+=+-4+%29
now find b
16-%28-4%29+=+b....eq.2
16%2B4+=+b
highlight%28b=20%29
so, your equation is: y=sqrt%28-4x+%2B20%29