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Write a quadratic function in standard form whose graph passes through the points (-5, 0), (-1, 0), and (-4, 3).​

User Mauridb
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1 Answer

6 votes

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.

User Lorde
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