Find the equation for the quadratic function whose x-intercept are -3 and 4 with the leading coefficient is 1

Answer 1

Hello Ace P.

To put it simply, the x-intercepts, or "zeros" are the values that, when you substitute them for x in the equation, make the function p(x) = 0 (or y = 0 on a graph). Also, remember that any time one of the values in a string of numbers is = 0, the product of the whole thing = 0.

So how to make the function = 0 when x = -3? How about a term like x + 3? If you substitute -3 for x, you get zero. Or, putting it another way, solve x + 3 = 0. Answer is x = -3. With similar logic, a term like x - 4 would equal 0 if x = 4.

So the quadratic function is (x + 3)(x - 4). You can see that a value of x = -3 or x = 4 would make one of the terms = 0, and therefore the product = 0. Now just use FOIL to multiply.

x² - 4x + 3x + 12 -> x² - x - 12 is the quadratic with x-intercepts at -3 and 4. Also, the leading coefficient is 1.

Let's check it. If x = -3, you get 9 + 3 - 12 (OK - that's = 0); If x = 4, you get 16 - 4 - 12 (OK, also = 0). So the quadratic function we created does have zeros, or x-intercepts, at -3 and 4.

Answer 2

y= (x+3)(x-4) = x^2 -x -12

y = f(x) = x^2 -x -12

f(-3) = (-3)^2 -(-3) -12 = 9+3 -12 = 0

f(4) = (4)^2 -(4) -12 = 16-4-12 = 0

take each x intercept and change its sign, then stick an x in front of it

so x=-3 becomes a factor (x+3)

x=-3

add 3 to both sides

x +3 = 0

(x+3) is one factor

then

x=4

subtract 4 from both sides

x-4 =0

(x-4) is the other factor

multiply the two factors together to get the quadratic function

(x+3)(x-4)

use FOIL

x^2 -4x +3x -12

= x^2 -x -12

or write in vertex form

it's an upward opening parabola

x^2 -x -12

x= 1/2 + or -(1/2)sqr(1+48)

x = 1/2 + or - 7/2 = -3 or 4

complete the square

x^2 -x + 1/4 = 12 + 1/4

(x-1/2)^2 = 49/4

x-1/2 = + or - 7/2

x =- 7/2+1/2 = -3 or 4 are x intercept

y = (1/2)^2 -1/2 -12 = 1/4 -1/2 -12 = -12 1/4

the vertex is (1/2, -49/4)

f(x) = (x-1/2)^2 - 49/4 is the quadratic in vertex form

x=1/2 is the line of symmetry

Answer 3

A quadratic function has the form y=ax² + bx + c where a b and c are numbers, y is the dependent variable (it depends on x to define what it is) and x is the independent variable (doesn't care about y).

At an x interecept y by definition is 0. So we know from the prompt that when x=-3 or x=4 y=0. We also know that the leading coefficient (here the leading coefficient in a, but in general the leading coefficient is the coefficient of the highest degree variable) is 1.

So we now have the following knowledge:

1) y=1x² +bx+c

2) 0=1(4)²+b(4)+c=16+4b+c

3) 0=1(-3)²+(-3)b+c=9-3b+c

So we have 2 equations (2 and 3 above) and 2 unknowns. We should be able to solve this equation. Let's solve one equation for c.

From 2) above: c=-16-4b

We can plug 2) into 3).

0=9-3b+(-16-4b)

0=9-3b-16-4b

0=-7-7b

7=-7b

-1=b

Plugging b into either 2) or 3) above

16+4b+c=0

16-4+c=0

12+c=0

c=-12

Now that we know c and b, we can plug these into the original equation and get our answer!

Original equation: y=x² + bx + c

y=x² - x -12

This is our solution. Let's check it! If this is indeed correct, then when y=0 x should equal -3 and 4

0=x² - x - 12

12=x² -x

12=x(x-1)

Let's plug in -3 and see what happens.

12=-3(-3-1)=-3(-4)=12 therefore 12=12 and -3 check passes

Let's plug in 4 and see what happens.

12=4(4-1)=4(3)=12 therefore 12=12 and 4 check passes.

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