Choose an equation in slope-intercept form for the line that passes through (-8,1) and is perpendicular to the y=2x-17.

Match the parabolas represented by the equations with their vertices. y = x2 + 6x + 8 y = 2x2 + 16x + 28 y = -x2 + 5x + 14 y = -

GaryK [48]

Consider all parabolas:

1.

$y = x^2 + 6x + 8,\\y=x^2+6x+9-9+8,\\y=(x^2+6x+9)-1,\\y=(x+3)^2-1.$

When x=-3, y=-1, then the point (-3,-1) is vertex of this first parabola.

2.

$y = 2x^2 + 16x + 28=2(x^2+8x+14),\\y=2(x^2+8x+16-16+14),\\y=2((x^2+8x+16)-16+14),\\y=2((x+4)^2-2)=2(x+4)^2-4.$

When x=-4, y=-4, then the point (-4,-4) is vertex of this second parabola.

3.

$y =-x^2 + 5x + 14=-(x^2-5x-14),\\y=-(x^2-5x+\dfrac{25}{4}-\dfrac{25}{4}-14),\\y=-((x^2-5x+\dfrac{25}{4})-\dfrac{25}{4}-14),\\y=-((x-\dfrac{5}{2})^2-\dfrac{81}{4})=-(x-\dfrac{5}{2})^2+\dfrac{81}{4}.$

When x=2.5, y=20.25, then the point (2.5,20.25) is vertex of this third parabola.

4.

$y =-x^2 + 7x + 7=-(x^2-7x-7),\\y=-(x^2-7x+\dfrac{49}{4}-\dfrac{49}{4}-7),\\y=-((x^2-7x+\dfrac{49}{4})-\dfrac{49}{4}-7),\\y=-((x-\dfrac{7}{2})^2-\dfrac{77}{4})=-(x-\dfrac{7}{2})^2+\dfrac{77}{4}.$

When x=3.5, y=19.25, then the point (3.5,19.25) is vertex of this fourth parabola.

5.

$y =2x^2 + 7x +5=2(x^2+\dfrac{7}{2}x+\dfrac{5}{2}),\\y=2(x^2+\dfrac{7}{2}x+\dfrac{49}{16}-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x^2+\dfrac{7}{2}x+\dfrac{49}{16})-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x+\dfrac{7}{4})^2-\dfrac{9}{16})=2(x+\dfrac{7}{4})^2-\dfrac{9}{8}.$

When x=-1.75, y=-1.125, then the point (-1.75,-1.125) is vertex of this fifth parabola.

6.

$y =-2x^2 + 8x +5=-2(x^2-4x-\dfrac{5}{2}),\\y=-2(x^2-4x+4-4-\dfrac{5}{2}),\\y=-2((x^2-4x+4)-4-\dfrac{5}{2}),\\y=-2((x-2)^2-\dfrac{13}{2})=-2(x-2)^2+13.$

When x=2, y=13, then the point (2,13) is vertex of this sixth parabola.

3 0

3 years ago

Find the equation of the line passing through the points (0,1) and (-2,0). write the equation in slope-intercept form.

Misha Larkins [42]

First of all, you have to calculate slope of that,

m = (y2-y1)/(x2-x1) = (1-0)/(0+2) = 1/2

now, equation would be y-y1 = m(x-x1)

y-1 = 1/2(x-0)

y-1 = x/2

y = x/2+ 1

5 0

3 years ago

The square root of pie must be at less ten numbers long

ZanzabumX [31]

The square root of pi is 1.77245385091.

4 0

2 years ago

Three vertices of the trapezoid are A(4d,4e), B(4f,4e), and C(4g,0). The fourth vertex lies on the origin. Find the midpoint of

satela [25.4K]

We have that

point C and point D have y = 0-----------> (the bottom of the trapezoid).

point A and point B have y = 4e ---------- > (the top of the trapezoid)

the y component of midpoint would be halfway between these lines
y = (4e+ 0)/2 = 2e.

<span>the x component of the midpoint of the midsegment would be halfway between the midpoint of AB and the midpoint of CD.

x component of midpoint of AB is (4d + 4f)/2.
x component of midpoint of CD is (4g + 0)/2 = 4g/2.
x component of a point between the two we just found is
[(4d + 4f)/2 + 4g/2]/2 = [(4d + 4f + 4g)/2]/2 = (4d + 4f + 4g)/4 = d + f + g.

</span>therefore

the midpoint of the midsegment is (d + f + g, 2e)

8 0

3 years ago

Model the pair of situations with exponential functions f and g. Find the approximate value of x that makes f(x)= g(x).

Law Incorporation [45]

I like the cat on your pfp LOL sorry i couldn’t help tho

6 0

2 years ago