What is the standard form of the equation of a quadratic function with roots of 4 and -1 that passes through [tex](1,-9)[/tex]?

A. [tex]y=1.5x^2-4.5x-6[/tex]
B. [tex]y=1.5x^2-4.5x+6[/tex]
C. [tex]y=-1.5x^2-4.5x-6[/tex]
D. [tex]y=-1.5x^2-4.5x+6[/tex]



Answer :

To find the standard form of the quadratic equation given its roots and a point it passes through, we follow these steps:

1. Identify the roots: The roots of the quadratic equation are 4 and -1.

2. Form the factored expression: Using the roots, the quadratic equation in its factored form can be written as:
[tex]\[ y = a(x - 4)(x + 1) \][/tex]
where [tex]\(a\)[/tex] is a constant to be determined.

3. Use the given point: The quadratic passes through the point [tex]\((1, -9)\)[/tex]. We will substitute [tex]\(x = 1\)[/tex] and [tex]\(y = -9\)[/tex] into the equation to find the value of [tex]\(a\)[/tex].

4. Substitute the point into the equation:
[tex]\[ -9 = a(1 - 4)(1 + 1) \][/tex]

5. Simplify and solve for [tex]\(a\)[/tex]:
[tex]\[ -9 = a(-3)(2) \][/tex]
[tex]\[ -9 = -6a \][/tex]
[tex]\[ a = \frac{-9}{-6} = \frac{3}{2} \][/tex]

6. Write the quadratic equation: Substitute [tex]\(a = \frac{3}{2}\)[/tex] back into the factored form:
[tex]\[ y = \frac{3}{2}(x - 4)(x + 1) \][/tex]

7. Expand to get the standard form:
[tex]\[ y = \frac{3}{2}(x^2 + x - 4x - 4) \][/tex]
[tex]\[ y = \frac{3}{2}(x^2 - 3x - 4) \][/tex]
[tex]\[ y = \frac{3}{2}x^2 - \frac{9}{2}x - 6 \][/tex]

Therefore, the standard form of the quadratic equation is:
[tex]\[ y = \frac{3}{2}x^2 - \frac{9}{2}x - 6 \][/tex]

Converting the coefficients to match the given options, we recognize that:

[tex]\(\frac{3}{2}\)[/tex] is equivalent to 1.5, and [tex]\(\frac{9}{2}\)[/tex] is equivalent to 4.5, we have:
[tex]\[ y = 1.5x^2 - 4.5x - 6 \][/tex]

So the correct answer is:
[tex]\[ y = 1.5x^2 - 4.5x - 6 \][/tex]