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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)?

y = 1.5x2 − 4.5x − 6
y = 1.5x2 − 4.5x + 6
y = −1.5x2 − 4.5x − 6
y = −1.5x2 − 4.5x + 6

User Brad Woods
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1 Answer

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Answer:

(a) y = 1.5x² -4.5x -6

Explanation:

You want the standard form of the quadratic with roots 4 and -1, and passing through the point (1, -9).

Coefficients

For roots p and q, the standard form can be found from the factored form:

(x -p)(x -q) = x² -(p+q)x +pq

For the given roots, the product pq will be ...

(4)(-1) = -4

The sum p+q is 4+(-1) = 3. The x-coefficient is the opposite of this.

The parent quadratic (before vertical scaling) will be ...

(x -4)(x +1) = x² -3x -4

The coefficients of the last two terms have the same sign. (Eliminates the 2nd and 4th answer choices.)

Leading coefficient

The given point has an x-value (1) between the given roots (-1, 4):

-1 < 1 < 4

The y-value at that point is negative. This means the vertex of the parabola will be below the x-axis, so the parabola opens upward and the leading coefficient is positive. (Eliminates the last two answer choices.)

The answer choices tell us the leading coefficient is 1.5, so the equation is ...

y = 1.5(x² -3x -4)

y = 1.5x² -4.5x -6 . . . . . . matches the first choice

__

Additional comment

We could find the value of the leading coefficient 'a' by evaluating ...

y = a(x -4)(x +1)

for x = 1. We would get ...

-9 = a(1 -4)(1 +1) = -6a ⇒ a = -9/-6 = 1.5

As we saw above, this isn't necessary. We only need to know that its sign is positive. The answer choices tell us the value.

If the x-value of the given point is not between the roots, then we know the sign of the y-value is the sign of the leading coefficient.

For multiple-choice questions, you only need to work enough of the problem to determine which answer choice is correct. You don't necessarily need to work the problem all the way to an answer.

What is the standard form of the equation of a quadratic function with roots of 4 and-example-1
User John Chuckran
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