PLEASE HELP! urgent math help!!?

Question

hi, i have a math test tomrrow and i cant figure out these questions on my review sheet.

1. determine the equations of the lines with slope-5 that intersect f(x)= -7x^2 -5x +4 once, twice or never.

2. also, how do you "solve the system" (i think they mean find the points of interesection). between y=4(x-2)(x+5) and y=-4x+44 ?

3. and this quick one. solve for x. 0= -5(x-3)^2

4. how do you know if two lines have more than one intersection?

i lkeep doing it and keep getting the wrong answers..




ANSWERS:
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1. y=-5x +k, k=4, one solution, k> 4 no solutions, k<4 two solutions.

2. (-7, 72) and (3,32)

3. 3 plus/minus squareroot of 2. approx 1.6 or 4.4

mcfate, oh yes there is! sry its actaully

0=-5(x-3)^2 +10

Answer 1

1) y = -5x + a
is a line sloping downward

y = f(x) = -7x^2 -5x + 4
is an upside-down parabola

These two will be equal when:
-5x + a = -7x^2 - 5x + 4
0 = -7x^2 + (4-a)
0 = 7x^2 - (4-a)

Therefore, x^2 = (4-a)/7
or x = +/- sqrt((4-a)/7)
- if 4 > a, there are two solutions
- if 4 = a, there is one solution: x = 0
- if 4 < a, there is no real solution

2) y = 4(x-2)(x+5) & y = -4x + 44

-4x + 44 = y = 4(x-2)(x+5)
-4x + 44 = 4(x^2 +3x -10) = 4x^2 +12x - 40
0 = 4x^2 + 16x - 84
0 = x^2 + 4x - 21
= (x-3)(x+7) so x = 3, -7
Then you find the values of y correspondingly, by either equation (they have to agree).

3) 0 = -5(x-3)^2
So
0 = (x-3)^2
=> x = 3
There must be a typographical error somewhere, if the answer is supposed to be sqrt(2).

4) Two lines
Either two lines intersect in only one point, or they are identical lines. That is basic geometry.

Answer 2

1. determine the equations of the lines with slope-5 that intersect f(x)= -7x^2 -5x +4 once, twice or never.

That's a parabola which opens downward. What you need to do is find the tangent with slope -5. All lines higher than that with the same slope would not intersect the parabola, and all lines lower than that would intersect it twice.

The slope f'(x) is:

f(x)= -7x^2 -5x +4
f'(x) = -14x - 5

Solve for slope -5:

f'(x) = -5
-14x - 5 = -5
-14x = 0
x = 0

The tangent with slope -5 touches the parabola at x=0:

f(x)= -7x^2 -5x +4
f(0) = 0 - 0 + 4
f(0) = 4

The point it touches is (0,4)

The equation for that line (slope -5, passing through (0,4) is:

y = -5x + 4

Lines with slope of -5 are of the form:

y = -5x + k

When k=4, the line is tangent to the parabola and intersects it once. When k<4, the line is lower and intersects the parabola twice. When k>4, the line is higher and does not intersect the parabola.

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2. also, how do you "solve the system" (i think they mean find the points of interesection). between y=4(x-2)(x+5) and y=-4x+44 ?

Since both equations equal y, you can set the two x-parts equal to each other:

4(x-2)(x+5) = -4x + 44
(x-2)(x+5) = -x + 11
x^2 + 3x - 10 = -x + 11
x^2 + 4x - 21 = 0
(x - 3)(x + 7) = 0

The solutions are x=3 and x=-7.

Now, plug back into either original equation to get the corresponding y-values:

y = -4x + 44
y = -4*3 + 44
y = -12 + 44
y = 32

One point is (3, 32)

y = -4x + 44
y = -4*(-7) + 44
y = -28 + 44
y = 16

The other point is (-7,16)

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3. and this quick one. solve for x. 0= -5(x-3)^2

The way this is written, x=3 is the only solution. Is it possible that there is a typo in the equation?

0=-5(x-3)^2 +10

0 = -5(x^2 -6x + 9) + 10
0 = -5x^2 + 30x - 45 + 10
5x^2 - 30x + 35 = 0
x^2 - 6x + 7 = 0

Use the quadratic equation:

x = ( -b +/- sqrt(b^2 - 4ac)) / 2a
x = (6 +/- sqrt((-6)^2 - 4*1*7)) / 2*1
x = 3 +/- sqrt(36 - 28)/2
x = 3 +/- sqrt(8)/2
x = 3 +/- sqrt(2)

Actually there's a shortcut you can use on this one:

0=-5(x-3)^2 +10
5(x-3)^2 = 10
(x-3)^2 = 2
x-3 = +/- sqrt(2)
x = 3 +/- sqrt(2)

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4. how do you know if two lines have more than one intersection?

The only way for two lines to have more than one intersection, is if they are the same line. The two equations would be the same, or a multiple of each other. (e.g., y = 2x + 4 vs. 3y = 6x + 12)

Answer 3

1) lines with a slope of -5 will all have a general equation of y = -5x +k

so we equate the two to find where these will intersect (ie have a point in common)

so

-5x +k = -7x^2 -5x +4

k = 4 - 7x^2

min value for x^2 = 0

therefore min value for k = 4 (- 7 x 0 ^2)

hence your guide answer begins to take shape

no solutions below 4 for k

2) equate the two equations and then solve to get the roots

so 4(x-2)(x+5) = 4x +44

divide by 4 throughout as a common factorand multiply out the brackets

x^2 +3x - 10 = x + 11

then bring everything onto one side leaving a 0 on the other side

x^2 +2x - 21 = 0

then factorise to get your values for x

then substitute into oneof the equations to get your corresponding value for y

3) -5 (x-3)^2 = 0

we can ignore the x -5 and the square sign when dealing with equating to zero

that leaves us with x-3 = 0

and hence the solution of x = 3

4. if you are talking poly nomials then i would

equate them

bring the factors all onto one side and equate to zero

then combine them

then factorise - this will give you all the combination points you need

if you merely want to know how many intersection points there are then a look at the highest power of the combined values would be able to tell you that at a glance (ie a x^4 power would tell you that you would probably get up to 4 spearate intersections)

Answer 4

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