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

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


Sagot :

To find the standard form of the quadratic equation with roots 4 and -1 that passes through the point \((1, -9)\), let's follow these steps:

### Step 1: Write the Quadratic Equation in Factored Form
Since the roots of the quadratic function are 4 and -1, we can write the quadratic function in its factored form as:
[tex]\[ y = a(x - 4)(x + 1) \][/tex]
Here, \(a\) is a constant that we need to determine.

### Step 2: Expand the Factored Form
Expand the equations \(y = a(x - 4)(x + 1)\):
[tex]\[ y = a(x^2 + x - 4x - 4) \][/tex]
[tex]\[ y = a(x^2 - 3x - 4) \][/tex]

### Step 3: Use the Given Point to Find \(a\)
We are given that the quadratic function passes through the point \((1, -9)\). This means when \(x = 1\), \(y = -9\).

Substitute \(x = 1\) and \(y = -9\) into the expanded equation:
[tex]\[ -9 = a(1^2 - 3(1) - 4) \][/tex]
Simplify inside the parentheses:
[tex]\[ -9 = a(1 - 3 - 4) \][/tex]
[tex]\[ -9 = a(-6) \][/tex]
Solve for \(a\):
[tex]\[ a = \frac{-9}{-6} \][/tex]
[tex]\[ a = \frac{3}{2} \][/tex]

### Step 4: Substitute \(a\) Back into the Quadratic Equation
Now that we have \(a = \frac{3}{2}\), we can substitute this back into the equation:
[tex]\[ y = \frac{3}{2}(x^2 - 3x - 4) \][/tex]

### Step 5: Write the Quadratic Equation in Standard Form
Distribute \(\frac{3}{2}\) to each term inside the parentheses:
[tex]\[ y = \frac{3}{2}x^2 - \frac{3}{2} \cdot 3x - \frac{3}{2} \cdot 4 \][/tex]
[tex]\[ y = \frac{3}{2}x^2 - \frac{9}{2}x - 6 \][/tex]

Now we can write the quadratic equation:
[tex]\[ y = 1.5x^2 - 4.5x - 6 \][/tex]

### Conclusion
The standard form of the quadratic equation that satisfies these conditions is:
[tex]\[ y = 1.5x^2 - 4.5x - 6 \][/tex]

Comparing this with the given options:
[tex]\[ \boxed{y = 1.5 x^2 - 4.5 x - 6} \][/tex]