Answer:
graph 1: f(x) = 3*x^2 - 5*x
graph 2: f(x) = 3*x^2 + 5
graph 3: f(x) = -3*x^2 - 5
Explanation:
This is ratter easy if we know some of the main rules for quadratic equations.
For a quadratic equation:
y = f(x) = a*x^2 + b*x + c
if a > 0, then the arms of the graph will open upwards.
if a < 0, then the arms of the graph open downwards.
With this in mind we can look at the graphs, only one of them have arms that open downwards, then the function that matches with this graph should be the only one with a leading coefficient smaller than zero, which is:
f(x) = -3*x^2 - 5
Now, when we have functions like:
y = f(x) = a*x^2 + c
c is the value of y at the vertex.
In graph 2 we can see that the arms open upwards, and that the y-value at the vertex is positive (it is above the horizontal axis)
Then the function of graph 2 should have a positive leading coefficient and a positive constant term.
The function is:
f(x) = 3*x^2 + 5
Then the remaining function is the one that corresponds to graph 1
f(x) = 3*x^2 - 5*x
Then we have:
graph 1: f(x) = 3*x^2 - 5*x
graph 2: f(x) = 3*x^2 + 5
graph 3: f(x) = -3*x^2 - 5