Answer:
f(x) = -x^2 - 6x + 13
Explanation:
To write a quadratic function in standard form that passes through the given points, we can use the fact that any quadratic function can be written in the form:
f(x) = a(x - h)^2 + k
where (h, k) is the vertex of the parabola and a is a coefficient that determines the shape of the parabola.
Since the graph passes through (-5, 0) and (-1, 0), we know that the vertex of the parabola is halfway between these two points, at x = -3. We can use this to find h:
h = -3
Now we need to find k and a. We know that the graph passes through (-4, 3), so we can substitute this point into the equation to get:
3 = a(-4 - (-3))^2 + k
3 = a(1)^2 + k
3 = a + k
We also know that the graph passes through the point (-5, 0), so we can substitute this point into the equation to get:
0 = a(-5 - (-3))^2 + k
0 = a(-2)^2 + k
0 = 4a + k
Now we have two equations with two unknowns (a and k), which we can solve simultaneously. Subtracting the second equation from the first, we get:
3 - 0 = a + k - (4a + k)
3 = -3a
Solving for a, we get:
a = -1
Substituting this value of a into either of the previous equations, we can solve for k:
3 = -1 + k
k = 4
Therefore, the quadratic function in standard form that passes through the given points is:
f(x) = -1(x + 3)^2 + 4
Expanding the square and simplifying, we get:
f(x) = -x^2 - 6x + 13
So the quadratic function in standard form is f(x) = -x^2 - 6x + 13.