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Which of the following equations have x-intercepts at (4,0) and (-7,0)? Oy = (2 - 4) - 28 Oy = (z- 4) (z +7) O y = x2 + 3z - 28 Oy = x - 4x +7

Which of the following equations have x-intercepts at (4,0) and (-7,0)? Oy = (2 - 4) - 28 Oy = (z- 4) (z +7) O y = x2 + 3z - 28 Oy = x - 4x +7

Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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### Question 12 Which of the following equations have x-intercepts at (4, 0) and (-7, 0)? 1. \( y = (x - 4)^2 - 28 \) 2. \( y = (x - 4)(x + 7) \) 3. \( y = x^2 + 3x - 28 \) 4. \( y = x^2 - 4x + 7 \) #### Explanation: For an equation to have x-intercepts at (4, 0) and (-7, 0), it means that when \( x = 4 \) and \( x = -7 \), the value of \( y \) will be 0. **Option 1: \( y = (x - 4)^2 - 28 \)** - This equation would form a parabola, but when \( x = 4 \), \( y \) does not equal 0. Thus, this cannot be the answer. **Option 2: \( y = (x - 4)(x + 7) \)** - Substitute \( x = 4 \): \[ y = (4 - 4)(4 + 7) = 0 \cdot 11 = 0 \] - Substitute \( x = -7 \): \[ y = (-7 - 4)(-7 + 7) = (-11) \cdot 0 = 0 \] - Therefore, this equation has the x-intercepts at both (4, 0) and (-7, 0). **Option 3: \( y = x^2 + 3x - 28 \)** - To find the x-intercepts, factorize the quadratic: \[ x^2 + 3x - 28 = (x - 4)(x + 7) \] - This is equivalent to option 2; therefore, it also has the x-intercepts at (4, 0) and (-7, 0). **Option 4: \( y = x^2 - 4x + 7 \)** - This cannot be factored into the form \( (x - 4)(x + 7) \). Thus, it does not have the x-intercepts at (4, 0) and (-7
Transcribed Image Text:### Question 12 Which of the following equations have x-intercepts at (4, 0) and (-7, 0)? 1. \( y = (x - 4)^2 - 28 \) 2. \( y = (x - 4)(x + 7) \) 3. \( y = x^2 + 3x - 28 \) 4. \( y = x^2 - 4x + 7 \) #### Explanation: For an equation to have x-intercepts at (4, 0) and (-7, 0), it means that when \( x = 4 \) and \( x = -7 \), the value of \( y \) will be 0. **Option 1: \( y = (x - 4)^2 - 28 \)** - This equation would form a parabola, but when \( x = 4 \), \( y \) does not equal 0. Thus, this cannot be the answer. **Option 2: \( y = (x - 4)(x + 7) \)** - Substitute \( x = 4 \): \[ y = (4 - 4)(4 + 7) = 0 \cdot 11 = 0 \] - Substitute \( x = -7 \): \[ y = (-7 - 4)(-7 + 7) = (-11) \cdot 0 = 0 \] - Therefore, this equation has the x-intercepts at both (4, 0) and (-7, 0). **Option 3: \( y = x^2 + 3x - 28 \)** - To find the x-intercepts, factorize the quadratic: \[ x^2 + 3x - 28 = (x - 4)(x + 7) \] - This is equivalent to option 2; therefore, it also has the x-intercepts at (4, 0) and (-7, 0). **Option 4: \( y = x^2 - 4x + 7 \)** - This cannot be factored into the form \( (x - 4)(x + 7) \). Thus, it does not have the x-intercepts at (4, 0) and (-7
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