Question

Find the formula, in standard form y = ax2 + bx + c, for a quadratic that has roots at x = -2 and x = 7,

and has leading coefficient of 1.

y =

Answer 1

Answer:

$y = x^2 - 5x - 14$

Step-by-step explanation:

Given

$x = -2$ and $x = 7$

Required

Determine the quadratic equation with a leading coefficient of 1

First, we equate both expression to 0

$x = -2$ and $x = 7$  becomes

$x + 2 = 0$ and $x - 7 = 0$

Multiply both expressions:

$(x + 2) * (x - 7) = 0 * 0$

$(x + 2) (x - 7) = 0 * 0$

Open Brackets

$x^2 - 7x + 2x - 14 = 0$

$x^2 - 5x - 14 = 0$

Hence, the quadratic equation is:

$x^2 - 5x - 14 = 0$

or

$y = x^2 - 5x - 14$

Answer 2

The formula, in standard form y = ax^2 + bx + c, for a quadratic that has roots at x = -2 and x = 7 is y = x² - 5x + 14

Given the standard form of quadratic equation expressed as $ax^2 + bx+c =0$

If a quadratic that has roots at x = -2 and x = 7, then their factors will be x+ 2 and x - 7.

Taking the product of the factors will give:

f(x) = (x+2)(x-7)

f(x) = x² - 7x + 2x - 14

f(x) = x² - 5x + 14

Hence the formula, in standard form y = ax^2 + bx + c, for a quadratic that has roots at x = -2 and x = 7 is y = x² - 5x + 14

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