To write a quadratic function in standard form that passes through (-4,0), (5,0), and (3,14), we can use the fact that a quadratic function has the form:
y = ax^2 + bx + c
where a, b, and c are constants. To find the values of these constants, we can substitute the coordinates of the three given points into the quadratic function and solve for a, b, and c. This gives us three equations:
0 = 16a - 4b + c
0 = 25a + 5b + c
14 = 9a + 3b + c
We can solve these equations using algebraic methods, such as elimination or substitution. One possible method is to subtract the first equation from the second equation and solve for b:
0 = 9a + 9b
b = -a
Substituting this into the third equation and simplifying, we get:
14 = 9a - 3a + c
14 = 6a + c
c = 14 - 6a
Substituting these values of b and c into the first equation and simplifying, we get:
0 = 16a + 4a(5) + (14 - 6a)
0 = 10a + 14
a = -7/5
Substituting this value of a into the expressions for b and c, we get:
b = 7/5
c = 98/5
Therefore, the quadratic function in standard form that passes through (-4,0), (5,0), and (3,14) is:
y = (-7/5)x^2 + (7/5)x + 98/5
or
y = -1.4x^2 + 1.4x + 19.6