Math Problem Statement

Part (a): Find a quadratic equation with a minimum point at (-2, 4). Part (b): Find a quadratic equation with a maximum point at (3, -5).

Solution

Let's solve the problem step by step.

Part (a): Find a quadratic equation with a minimum point at (2,4)(-2, 4).

The general form of a quadratic function is: y=a(xh)2+ky = a(x - h)^2 + k Where (h,k)(h, k) is the vertex of the parabola.

Given that the minimum point is at (2,4)(-2, 4), this point represents the vertex. So we substitute h=2h = -2 and k=4k = 4:

y=a(x+2)2+4y = a(x + 2)^2 + 4

Now, to fully define the equation, we need the value of aa. Since the problem doesn't give any other specific information, we can assume a=1a = 1 (a common assumption for a simple quadratic equation). Therefore, the quadratic equation becomes:

y=(x+2)2+4y = (x + 2)^2 + 4

Expanding this, we get: y=(x2+4x+4)+4y = (x^2 + 4x + 4) + 4 y=x2+4x+8y = x^2 + 4x + 8

So, the quadratic equation is: y=x2+4x+8y = x^2 + 4x + 8

Part (b): Find a quadratic equation with a maximum point at (3,5)(3, -5).

Similarly, the vertex form of the quadratic equation is: y=a(xh)2+ky = a(x - h)^2 + k For a maximum point at (3,5)(3, -5), we substitute h=3h = 3 and k=5k = -5:

y=a(x3)25y = a(x - 3)^2 - 5

Since this is a maximum point, the parabola must open downwards, meaning aa is negative. For simplicity, we assume a=1a = -1. Therefore, the equation becomes:

y=(x3)25y = -(x - 3)^2 - 5

Expanding this, we get: y=(x26x+9)5y = -(x^2 - 6x + 9) - 5 y=x2+6x95y = -x^2 + 6x - 9 - 5 y=x2+6x14y = -x^2 + 6x - 14

So, the quadratic equation is: y=x2+6x14y = -x^2 + 6x - 14

Summary:

  • (a) The quadratic equation with a minimum point at (2,4)(-2, 4) is y=x2+4x+8y = x^2 + 4x + 8.
  • (b) The quadratic equation with a maximum point at (3,5)(3, -5) is y=x2+6x14y = -x^2 + 6x - 14.

Let me know if you'd like any further explanation or details!


Here are 5 related questions:

  1. How do you determine the direction a parabola opens?
  2. What happens to the quadratic equation if aa is negative versus positive?
  3. How can you find the axis of symmetry for a quadratic equation?
  4. What other methods can you use to find a quadratic equation from given points?
  5. How do you find the roots of a quadratic equation if the vertex is known?

Tip: The vertex form of a quadratic function y=a(xh)2+ky = a(x - h)^2 + k is especially useful when dealing with maximum or minimum points.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Functions
Vertex Form

Formulas

y = a(x - h)^2 + k
y = ax^2 + bx + c

Theorems

Vertex Theorem for Quadratics

Suitable Grade Level

Grades 9-12