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3 years ago

HELP URGENT - put 27 points on question please help!

Mathematics

1 answer:

Olenka [21]3 years ago

3 0

Answer:

f(x) = 3x² - 12x -135

Step-by-step explanation:

standard form of a quadratic equation is

y = Ax² + Bx + C

You are given 3 solutions for X and Y, i.e( x=-5, y = 0), (x = 9,y = 0) and (x = 8,y = -39)

Substitute each of this equations into the quadratic equation to obtain a system of 3 equations

For ( x=-5, y = 0), 25A - 5B + C = 0 ---------- eq (1)

For ( x= 9, y = 0), 81A + 9B + C = 0 ---------- eq (2)

For ( x= 8, y = -39), 64A + 8B + C = -39 ---------- eq (3)

You have 3 equations and 3 unknowns. Solving this system of 3 equations will give:

A = 3, B = -12, c = -135

Hence the quadratic equation is

y = 3x² - 12x -135

or in function form:

f(x) = 3x² - 12x -135

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Read 2 more answers

Help me please i don't know what to do

astra-53 [7]

Answer:

a. a = 1, b = -5, c = -14

b. a = 1, b = -6, c = 9

c. a = -1, b = -1, c = -3

d. a = 1, b = 0, c = -1

e. a = 1, b = 0, c = -3

Step-by-step explanation:

a. x-ints at 7 and -2

this means that our quadratic equation must factor to:

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

FOIL:

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

Simplify:

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

a = 1, b = -5, c = -14

b. one x-int at 3

this means that the equation will factor to:

$(x-3)^2=0$

FOIL:

$x^2 - 3x - 3x +9 = 0$

Simplify:

$x^2 - 6x + 9 = 0$

a = 1, b = -6, c = 9

c. no x-int and negative y must be less than 0

This means that our vertex must be below the x-axis and our parabola must point down

There are many equations for this, but one could be:

$-x^2-x-3=0$

a = -1, b = -1, c = -3

d. one positive x-int, one negative x-int

We can use any x-intercepts, so let's just use -1 and 1

The equation will factor to:

$(x+1)(x-1)=0$

This is a perfect square

FOIL:

$x^2-1=0$

a = 1, b = 0, c = -1

e. x-int at $\sqrt{3} , -\sqrt{3}$

our equation will factor to:

$(x+\sqrt{3} )(x-\sqrt{3)} =0$

This is also a perfect square

FOIL and you will get:

$x^2 - 3 = 0$

a = 1, b = 0, c = -3

4 0

2 years ago