Question

Can someone help me please​

Answer 1

The quadratic function, with intercepts at -5, 1, and 5, is y = 0 in intercept form. In standard form, after expanding, it is y =
`(4/5)x^3 - (4/5)x^2 - (76/5)x + 5.`

a. Intercept Form:

The quadratic function in intercept form is given by y = a(x - p)(x - q), where p and q are the x-intercepts.

1. X-intercepts (roots): Set y = 0 to find intercepts. Roots are x = -5, 1, 5.

2. Equation:

y = a(x + 5)(x - 1)(x - 5)

3. Substitute a point (e.g., x = 3, y = 0):

`\[ 0 = a(8)(4)(-2) \implies a = 0 \]`

4. Final Equation:

y = 0 (This quadratic has roots but no vertical stretch; it's a straight line)

b. Standard Form:

The standard form of a quadratic is
`\(y = ax^2 + bx + c\).`

1. Expand Intercept Form:

`\[ y = a(x + 5)(x - 1)(x - 5) \]`

`\[ y = a(x^3 - x^2 - 19x + 25) \]`

2. Substitute another point (e.g., x = -6, y = 9):

`\[ 9 = a(-216 + 36 + 114 + 25) \implies a = (4)/(5) \]`

3. Final Equation:

`\[ y = (4)/(5)x^3 - (4)/(5)x^2 - (76)/(5)x + 5 \]`