**Problem: Writing the Equation of a Parabola** Write an equation for a parabola with x-intercepts at (-3, 0) and (5, 0) which passes through the point (4, -28). **Instructions:** 1. Use the intercept form of a quadratic equation: \( y = a(x - p)(x - q) \). 2. Identify the x-intercepts as \( p = -3 \) and \( q = 5 \). 3. Substitute the point (4, -28) into the equation to solve for \( a \). **Steps:** 1. Begin with the general equation: \( y = a(x + 3)(x - 5) \). 2. Substitute the point (4, -28) into the equation: \[ -28 = a(4 + 3)(4 - 5) \] 3. Simplify and solve for \( a \): \[ -28 = a(7)(-1) \] \[ -28 = -7a \] \[ a = 4 \] **Final Equation:** The equation of the parabola is: \[ y = 4(x + 3)(x - 5) \]

Problem: Writing the Equation of a Parabola Write an equation for a parabola with x-intercepts at (-3, 0) and (5, 0) which passes through the point (4, -28). Instructions: 1. Use the intercept form of a quadratic equation: \( y = a(x - p)(x - q) \). 2. Identify the x-intercepts as \( p = -3 \) and \( q = 5 \). 3. Substitute the point (4, -28) into the equation to solve for \( a \). Steps: 1. Begin with the general equation: \( y = a(x + 3)(x - 5) \). 2. Substitute the point (4, -28) into the equation: \[ -28 = a(4 + 3)(4 - 5) \] 3. Simplify and solve for \( a \): \[ -28 = a(7)(-1) \] \[ -28 = -7a \] \[ a = 4 \] Final Equation: The equation of the parabola is: \[ y = 4(x + 3)(x - 5) \]

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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Question

Write an equation for a parabola with X intercepts (-3,0) and (5,0)which passes through the point (4 -28)

**Problem: Writing the Equation of a Parabola**

Write an equation for a parabola with x-intercepts at (-3, 0) and (5, 0) which passes through the point (4, -28).

**Instructions:**

1. Use the intercept form of a quadratic equation: \( y = a(x - p)(x - q) \).
2. Identify the x-intercepts as \( p = -3 \) and \( q = 5 \).
3. Substitute the point (4, -28) into the equation to solve for \( a \).

**Steps:**

1. Begin with the general equation: \( y = a(x + 3)(x - 5) \).
2. Substitute the point (4, -28) into the equation:
\[
-28 = a(4 + 3)(4 - 5)
\]
3. Simplify and solve for \( a \):
\[
-28 = a(7)(-1)
\]
\[
-28 = -7a
\]
\[
a = 4
\]

**Final Equation:**

The equation of the parabola is:
\[
y = 4(x + 3)(x - 5)
\]

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