I already drew the graph!

Condition 1 : The graph in the image is a parabola curve, therefore the function should be quadratic.

Therefore, functions 1 and 5 do not satisfy this.

From here, all explanations will be based on the general form of the quadratic function $f(x) = ax^2 + bx + c$

Condition 2 : It is a u-shaped parabola, which means that the coefficient of $x^2$ is positive, that is, $a>0$ .

Therefore, function 7 does not satisfy this too.

Condition 3 : We know that the $y$ -intercept is below the $x$ -axis, therefore the $y$ -intercept is negative, that is, $c<0$

Function 2 does not satisfy this condition.

We are now down to 3 possibilities left.

Condition 4 : Next, we need to ensure that the function has 2 different real roots, which translates to $b^2 - 4ac > 0$

Function 3: $b^2 -4ac = (-4)^2 - 4(1)(-5) = 16 + 20 = 36 > 0$

Function 4: $b^2 -4ac = (4)^2 - 4(1)(-5) = 16 + 20 = 36 > 0$

Function 6: $b^2 -4ac = (-8)^2 - 4(4)(-21) = 64 + 336 = 400 > 0$

Fun fact: The determinants for these 3 functions are all perfect squares, therefore their roots are all rational numbers

Condition 5 : Lastly, the minimum point is located in quadrant 4. The $x$ -value of the minimum point is positive $(x>0)$ , and the $y$ -value of the minimum point is negative $(y<0)$ . We need to check the minimum points of each function. Here, I will demonstrate 3 different methods to find the minimum value of a quadratic function.

Function 3 (Method 1):

For quadratic functions, the minimum point is defined as $\left(-\dfrac{b}{2a},f\left(-\dfrac{b}{2a}\right)\right)$

$-\dfrac{b}{2a} = -\dfrac{-4}{2(1)} = 2 > 0$

$f\left(-\dfrac{b}{2a}\right) = f(2) = 2^2 - 4(2) - 5 = -9 < 0$

The minimum point is $(2,-9)$ , and it satisfies Condition 5 .

Function 4 (Method 2):

For quadratic functions, you can complete the square and write it in the form of $f(x) = a(x-p)^2 + q$ . The minimum/maximum point is then defined as $(p,q)$ .

$x^2+4x-5\\ =x^2+4x+4-4-5\\ =(x+2)^2 - 9$

The minimum point is $(-2,-9)$ , and the $x$ -coordinate does not satisfy Condition 5 .

Function 6 (Method 3):

At minimum points, the first derivative, $f'(x) = 0$ and the second derivative, $f''(x)>0$

$f(x) = 4x^2 - 8x - 21\\ f'(x) = 8x - 8\\ f''(x) = 8 > 0$

We have shown that the critical point is a minimum point, now we need to find the coordinates of the minimum point.

$f'(x) = 0\\ 8x-8=0\\ 8x=8 \implies x=1\\ f(1) = 4(1^2) - 8(1) - 21 = -25$

The minimum point is $(1,-25)$ , and it satisfies Condition 5 .

Therefore, function 3, $f(x) = x^2 - 4x - 5$ and function 6, $f(x) = 4x^2 - 8x - 21$ can have the graph shown in the image in the question.

The answer is $\boxed{2}$