IDNLearn.com is the perfect place to get detailed and accurate answers to your questions. Our platform is designed to provide reliable and thorough answers to all your questions, no matter the topic.

What is the standard form of the equation of a quadratic function with roots of 2 and -5 that passes through [tex](1, -3)[/tex]?

A. [tex]y = -0.5x^2 + 1.5x - 5[/tex]
B. [tex]y = -0.5x^2 + 1.5x + 5[/tex]
C. [tex]y = 0.5x^2 + 1.5x - 5[/tex]
D. [tex]y = 0.5x^2 + 1.5x + 5[/tex]


Sagot :

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

1. Identify the given roots: The roots of the quadratic function are [tex]\( r_1 = 2 \)[/tex] and [tex]\( r_2 = -5 \)[/tex].
2. Form the quadratic equation: The quadratic can be expressed as:
[tex]\[ y = a (x - r_1)(x - r_2) \][/tex]
Substituting the given roots, we have:
[tex]\[ y = a (x - 2)(x + 5) \][/tex]

3. Convert to standard form:
Expand the expression inside the parentheses:
[tex]\[ y = a (x^2 + 5x - 2x - 10) \][/tex]
Simplify inside the parentheses:
[tex]\[ y = a (x^2 + 3x - 10) \][/tex]

4. Use the given point [tex]\((1, -3)\)[/tex] to find [tex]\(a\)[/tex]:
Substitute [tex]\(x = 1\)[/tex] and [tex]\(y = -3\)[/tex] into the equation:
[tex]\[ -3 = a (1^2 + 3(1) - 10) \][/tex]
Simplify the equation:
[tex]\[ -3 = a (1 + 3 - 10) \][/tex]
[tex]\[ -3 = a (-6) \][/tex]
Solving for [tex]\(a\)[/tex]:
[tex]\[ a = \frac{-3}{-6} = 0.5 \][/tex]

5. Form the final equation:
Substitute [tex]\(a = 0.5\)[/tex] back into the expanded form of the equation:
[tex]\[ y = 0.5 (x^2 + 3x - 10) \][/tex]

6. Distribute [tex]\(0.5\)[/tex]:
[tex]\[ y = 0.5x^2 + 1.5x - 5 \][/tex]

Therefore, the standard form of the quadratic function is:
[tex]\[ y = 0.5x^2 + 1.5x - 5 \][/tex]

Among the given options, the correct equation is:

[tex]\[ y = 0.5x^2 + 1.5x - 5 \][/tex]