SOLUTION: Find the equation of the quadratic function f whose graph is shown below. Ponts on the graph are (-3,-1)(-2,-4) f(x)=

Algebra ->  Graphs -> SOLUTION: Find the equation of the quadratic function f whose graph is shown below. Ponts on the graph are (-3,-1)(-2,-4) f(x)=       Log On


   



Question 399724: Find the equation of the quadratic function f whose graph is shown below.
Ponts on the graph are (-3,-1)(-2,-4)
f(x)=

Found 3 solutions by MathLover1, Edwin McCravy, richard1234:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

given points:
(-3,-1)(-2,-4)

Solved by pluggable solver: FIND EQUATION of straight line given 2 points
hahaWe are trying to find equation of form y=ax+b, where a is slope, and b is intercept, which passes through points (x1, y1) = (-3, -1) and (x2, y2) = (-2, -4).
Slope a is a+=+%28y%5B2%5D-y%5B1%5D%29%2F%28x%5B2%5D-x%5B1%5D%29+=+%28-4--1%29%2F%28-2--3%29+=+-3.
Intercept is found from equation a%2Ax%5B1%5D%2Bb+=+y%5B1%5D, or -3%2A-3+%2Bb+=+-10. From that,
intercept b is b=y%5B1%5D-a%2Ax%5B1%5D, or b=-1--3%2A-3+=+-10.

y=(-3)x + (-10)

Your graph:





f%28x%29=+%28-3%29x+-10

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!


The other tutor gave a linear, not quadratic solution, which is
not what you want.

Since we cannot see the graph that you can see, we cannot complete
your problem, for just having two point is not enough to determine
a quadratic function.  But I'll do something. 

The general equation of a quadratic function is

       f(x) = a*x² + b*x + c

Since there are three unknown constants a, b, and c, the two points you gave,
(-3,-1), (-2,-4), are not enough, so I will assume that you can look and see
a third point on the graph that you didn't tell us about. So I will arbitrarily
assume the y-intercept is (0,-4).  So I'll assume the graph you see that we
cannot see is this:

Here's the graph I'm assuming:



Substitute the point (-3,-1)

       f(x) = a*x² + b*x + c
         -1 = a*(-3)² + b*(-3) + c
         -1 = 9a - 3b + c
9a - 3b + c = -1 

Substitute the point (-2,-4)

       f(x) = a*x² + b*x + c
         -4 = a*(-2)² + b*(-2) + c
         -4 = 4a - 2b + c
4a - 2b + c = -4

Substitute the point, y-intercept, that I chose arbitrarily (0,-4)

       f(x) = a*x² + b*x + c
         -4 = a*(0)² + b*(0) + c
         -4 = 0a + 0b + c
          c = -4

So we have this system of equations:

          9a - 3b + c = -1
          4a - 2b + c = -4
                    c = -4

Solve that system and get a=1, b=2, c=-4

So the function is

f(x) = ax² + bx + c

f(x) = x² + 2x - 4

Edwin

Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
-1+=+a%28-3%29%5E2+%2B+b%28-3%29+%2B+c --> -1+=+9a+-+3b+%2B+c

-4+=+a%28-2%29%5E2+%2B+b%28-2%29+%2B+c --> -4+=+4a+-+2b+%2B+c

The quadratic equation cannot be uniquely determined since there are three variables and two equations.