Finding Quadratic Functions: Vertex & Point Guide
Hey math enthusiasts! Let's dive into the world of quadratic functions. Today, we're going to learn how to write a quadratic function in standard form when we're given some key information: the vertex and a point on the parabola. It's like having two puzzle pieces and figuring out the rest of the picture! We'll go through several examples step-by-step, making sure you grasp the concept and feel confident in solving these types of problems. So, grab your pencils, and let's get started. We will explore each problem carefully and use the standard form to express each quadratic function.
Understanding the Standard Form and Vertex Form
Before we jump into the examples, let's refresh our memory about the standard form and vertex form of a quadratic function. Understanding these forms is crucial for our task. The standard form of a quadratic function is written as:
- f(x) = ax² + bx + c
Where a, b, and c are constants, and a is not equal to zero. This form is great for identifying the y-intercept (the point where the parabola crosses the y-axis), which is the value of c.
However, for our purpose – finding a quadratic function when we know the vertex – the vertex form is much more useful. The vertex form of a quadratic function is given by:
- f(x) = a(x - h)² + k
Here, (h, k) represents the vertex of the parabola, and a is the same constant as in the standard form. The vertex form immediately gives us the coordinates of the vertex, which is super convenient when we're given the vertex as part of the problem. Our main goal will be to find the value of a. Once we know a, we can convert the vertex form back to standard form if needed (although that's not strictly required for these problems). So, keep in mind: Vertex form is the go-to when you're given the vertex!
Example 1: Vertex (0, 0), Point (2, 4)
Alright, let's tackle our first problem. We're given that the vertex is at the origin, (0, 0), and the parabola passes through the point (2, 4). Here's how we'll do it:
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Start with the Vertex Form: Since we have the vertex, we begin with the vertex form:
f(x) = a(x - h)² + k -
Plug in the Vertex: Our vertex is (0, 0), so h = 0 and k = 0. Substitute these values into the vertex form:
f(x) = a(x - 0)² + 0
which simplifies to: f(x) = ax² -
Use the Point (2, 4): We know the parabola passes through (2, 4). This means when x = 2, f(x) = 4. Substitute these values into our equation:
4 = a(2)²
4 = 4a -
Solve for a: Divide both sides by 4:
a = 1 -
Write the Equation: Now that we know a = 1, plug it back into our simplified vertex form:
f(x) = 1x² or simply
f(x) = x²
So, the quadratic function in standard form is f(x) = x². Easy peasy, right? The point helps us pinpoint the stretch or compression of the parabola.
Example 2: Vertex (2, 1), Point (4, 5)
Let's keep the momentum going! This time, our vertex is (2, 1), and the parabola passes through (4, 5). Let's repeat the steps:
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Vertex Form: Start with f(x) = a(x - h)² + k
-
Plug in the Vertex: Our vertex is (2, 1), so h = 2 and k = 1. Substitute:
f(x) = a(x - 2)² + 1 -
Use the Point (4, 5): When x = 4, f(x) = 5. Substitute:
5 = a(4 - 2)² + 1
5 = a(2)² + 1
5 = 4a + 1 -
Solve for a: Subtract 1 from both sides:
4 = 4a
Divide by 4: a = 1 -
Write the Equation: Substitute a = 1 back into the equation:
f(x) = 1(x - 2)² + 1 or simply
f(x) = (x - 2)² + 1
If we want to expand this into standard form, we would get:
f(x) = x² - 4x + 4 + 1
f(x) = x² - 4x + 5
Thus the quadratic function in standard form is f(x) = (x - 2)² + 1, or f(x) = x² - 4x + 5.
Example 3: Vertex (2, -4), Point (0, 0)
Alright, let's keep going. Now our vertex is (2, -4), and we have a point (0, 0). This problem will be quite interesting:
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Vertex Form: Start with f(x) = a(x - h)² + k
-
Plug in the Vertex: Our vertex is (2, -4), so h = 2 and k = -4. Substitute:
f(x) = a(x - 2)² - 4 -
Use the Point (0, 0): When x = 0, f(x) = 0. Substitute:
0 = a(0 - 2)² - 4
0 = a(-2)² - 4
0 = 4a - 4 -
Solve for a: Add 4 to both sides:
4 = 4a
Divide by 4: a = 1 -
Write the Equation: Substitute a = 1 back into the equation:
f(x) = 1(x - 2)² - 4 or simply
f(x) = (x - 2)² - 4
If we want to expand this into standard form, we would get:
f(x) = x² - 4x + 4 - 4
f(x) = x² - 4x
Thus the quadratic function in standard form is f(x) = (x - 2)² - 4, or f(x) = x² - 4x.
Example 4: Vertex (-4, -2), Point (-3, -1)
Let's tackle this example where the vertex has negative coordinates! Our vertex is (-4, -2), and we have the point (-3, -1). Let's walk through it:
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Vertex Form: Start with f(x) = a(x - h)² + k
-
Plug in the Vertex: Our vertex is (-4, -2), so h = -4 and k = -2. Substitute:
f(x) = a(x - (-4))² - 2 f(x) = a(x + 4)² - 2 -
Use the Point (-3, -1): When x = -3, f(x) = -1. Substitute:
-1 = a(-3 + 4)² - 2
-1 = a(1)² - 2
-1 = a - 2 -
Solve for a: Add 2 to both sides:
1 = a -
Write the Equation: Substitute a = 1 back into the equation:
f(x) = 1(x + 4)² - 2 or simply
f(x) = (x + 4)² - 2
If we want to expand this into standard form, we would get:
f(x) = x² + 8x + 16 - 2
f(x) = x² + 8x + 14
Thus the quadratic function in standard form is f(x) = (x + 4)² - 2, or f(x) = x² + 8x + 14.
Example 5: Vertex (3, -2), Point (5, 2)
Last one, guys! Our vertex is (3, -2), and the parabola passes through (5, 2). Let's find the quadratic function:
-
Vertex Form: Start with f(x) = a(x - h)² + k
-
Plug in the Vertex: Our vertex is (3, -2), so h = 3 and k = -2. Substitute:
f(x) = a(x - 3)² - 2 -
Use the Point (5, 2): When x = 5, f(x) = 2. Substitute:
2 = a(5 - 3)² - 2
2 = a(2)² - 2
2 = 4a - 2 -
Solve for a: Add 2 to both sides:
4 = 4a
Divide by 4: a = 1 -
Write the Equation: Substitute a = 1 back into the equation:
f(x) = 1(x - 3)² - 2 or simply
f(x) = (x - 3)² - 2
If we want to expand this into standard form, we would get:
f(x) = x² - 6x + 9 - 2
f(x) = x² - 6x + 7
Thus the quadratic function in standard form is f(x) = (x - 3)² - 2, or f(x) = x² - 6x + 7.
Conclusion: You Got This!
And there you have it! We've successfully written quadratic functions in standard form given the vertex and a point for several examples. Remember these key steps:
- Start with the vertex form: f(x) = a(x - h)² + k.
- Substitute the vertex (h, k).
- Substitute the x and f(x) values from the given point.
- Solve for a.
- Write the final equation.
Keep practicing, and you'll become a pro at this. Keep in mind that we can easily convert from vertex form to standard form using algebraic expansion. Keep up the great work, and don't hesitate to ask if you have any questions! Happy calculating!
