Math Problem Statement

Please solve this

Solution

To solve each part of the problem, we need to use the given quadratic equation y=a(xr)(xs)y = a(x - r)(x - s) and apply the conditions provided in each subpart. Let's go through each one step-by-step:

Part (a)

Given: The parabola has zeros at (4,0)(4, 0) and (2,0)(2, 0) and a y-intercept at (0,1)(0, 1).

Equation form: y=a(x4)(x2)y = a(x - 4)(x - 2).

To find aa, we use the y-intercept (0,1)(0, 1): 1=a(04)(02)1 = a(0 - 4)(0 - 2) 1=a(4)(2)1 = a \cdot (-4) \cdot (-2) 1=8a1 = 8a a=18a = \frac{1}{8}

Part (b)

Given: The parabola has x-intercepts at (4,0)(4, 0) and (2,0)(-2, 0) and a y-intercept at (0,1)(0, -1).

Equation form: y=a(x4)(x+2)y = a(x - 4)(x + 2).

To find aa, we use the y-intercept (0,1)(0, -1): 1=a(04)(0+2)-1 = a(0 - 4)(0 + 2) 1=a(4)2-1 = a \cdot (-4) \cdot 2 1=8a-1 = -8a a=18a = \frac{1}{8}

Part (c)

Given: The parabola has zeros at (5,0)(5, 0) and (0,0)(0, 0) and a minimum value of 10-10.

Equation form: y=a(x5)xy = a(x - 5)x.

The vertex (minimum) of the parabola is at x=5+02=2.5x = \frac{5 + 0}{2} = 2.5.

Using the minimum value 10-10 at x=2.5x = 2.5: 10=a(2.55)(2.5)-10 = a(2.5 - 5)(2.5) 10=a(2.5)(2.5)-10 = a(-2.5)(2.5) 10=a(6.25)-10 = a(-6.25) a=106.25=106.25=4025=85a = \frac{-10}{-6.25} = \frac{10}{6.25} = \frac{40}{25} = \frac{8}{5}

Part (d)

Given: The parabola has x-intercepts at (5,0)(5, 0) and (3,0)(-3, 0) and a maximum value of 6.

Equation form: y=a(x5)(x+3)y = a(x - 5)(x + 3).

The vertex (maximum) of the parabola is at x=5+(3)2=1x = \frac{5 + (-3)}{2} = 1.

Using the maximum value 66 at x=1x = 1: 6=a(15)(1+3)6 = a(1 - 5)(1 + 3) 6=a(4)(4)6 = a(-4)(4) 6=a(16)6 = a(-16) a=616=38a = \frac{6}{-16} = -\frac{3}{8}

Part (e)

Given: The parabola has its vertex at (5,0)(5, 0) and a y-intercept at (0,10)(0, -10).

The vertex form of a parabola is y=a(xh)2+ky = a(x - h)^2 + k, where (h,k)(h, k) is the vertex. Here, the vertex is at (5,0)(5, 0): y=a(x5)2y = a(x - 5)^2.

To find aa, we use the y-intercept (0,10)(0, -10): 10=a(05)2-10 = a(0 - 5)^2 10=a(25)-10 = a(25) a=1025=25a = \frac{-10}{25} = -\frac{2}{5}

These are the values of aa for each condition:

  • (a) a=18a = \frac{1}{8}
  • (b) a=18a = \frac{1}{8}
  • (c) a=85a = \frac{8}{5}
  • (d) a=38a = -\frac{3}{8}
  • (e) a=25a = -\frac{2}{5}

Do you want details or have any questions?

Here are some related questions for further practice:

  1. How do you find the vertex of a parabola given its standard form equation?
  2. What are the steps to convert a quadratic equation from standard form to vertex form?
  3. How do you determine the direction in which a parabola opens?
  4. What is the significance of the coefficient aa in a quadratic equation?
  5. How can you find the axis of symmetry for a parabola given its equation in standard form?
  6. What methods can be used to find the roots of a quadratic equation?
  7. How does the value of aa affect the width of the parabola?
  8. What is the relationship between the discriminant and the nature of the roots of a quadratic equation?

Tip: When solving quadratic equations, always check the given conditions (like intercepts, vertex, etc.) to determine the correct form of the equation and values of the coefficients.

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Math Problem Analysis

Mathematical Concepts

Quadratic Equations
Vertex Form
Intercepts

Formulas

Quadratic equation for parabolas

Theorems

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Suitable Grade Level

High School