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jeyben [28]
3 years ago
9

2. y = k2 - 13k +42​

Mathematics
1 answer:
bekas [8.4K]3 years ago
6 0

For this case we must solve the following quadratic equation:

y = k ^ 2-13k + 42

With y = 0 we have:

k ^ 2-13k + 42 = 0

The roots will be given by:

k = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}

Where:

a = 1\\b = -13\\c = 42

Substituting:

k = \frac {- (- 13) \pm \sqrt {(- 13) ^ 2-4 (1) (42)}} {2 (1)}\\k = \frac {13 \pm \sqrt {169-168}} {2}\\k = \frac {13 \pm \sqrt {1}} {2}\\k = \frac {13 \pm1} {2}

Thus, we have two roots:

k_ {1} = \frac {13 + 1} {2} = \frac {14} {2} = 7\\k_ {2} = \frac {13-1} {2} = \frac {12} {2} = 6

Answer:

k_ {1} = 7\\k_ {2} = 6

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dimaraw [331]

1. f(x)=x²+10x+16

Use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=1(As, a>0 the parabola is open upward), b=10. by putting the values.

-b/2a = -10/2(1) = -5

f(-b/2a)= f(-5)= (-5)²+10(-5)+16= -9

So, Vertex = (-5, -9)

Now, find y- intercept put x=0 in the above equation. f(0)= 0+0+16, we get point (0,16).

Now find x-intercept put y=0 in the above equation. 0= x²+10x+16

x²+10x+16=0 ⇒x²+8x+2x+16=0 ⇒x(x+8)+2(x+8)=0 ⇒(x+8)(x+2)=0 ⇒x=-8 , x=-2

From vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. The graph is attached below.

2. f(x)=−(x−3)(x+1)

By multiplying the factors, the general form is f(x)= -x²+2x+3.

Use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(As, a<0 the parabola is open downward), b=2. by putting the values.

-b/2a = -2/2(-1) = 1

f(-b/2a)= f(1)=-(1)²+2(1)+3= 4

So, Vertex = (1, 4)

Now, find y- intercept put x=0 in the above equation. f(0)= 0+0+3, we get point (0, 3).

Now find x-intercept put y=0 in the above equation. 0= -x²+2x+3.

-x²+2x+3=0 the factor form is already given in the question so, ⇒-(x-3)(x+1)=0 ⇒x=3 , x=-1

From vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. The graph is attached below.

3. f(x)= −x²+4

Use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(As, a<0 the parabola is open downward), b=0. by putting the values.

-b/2a = -0/2(-1) = 0

f(-b/2a)= f(0)= −(0)²+4 =4

So, Vertex = (0, 4)

Now, find y- intercept put x=0 in the above equation. f(0)= −(0)²+4, we get point (0, 4).

Now find x-intercept put y=0 in the above equation. 0= −x²+4

−x²+4=0 ⇒-(x²-4)=0 ⇒ -(x-2)(x+2)=0 ⇒x=2 , x=-2

From vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. The graph is attached below.

4. f(x)=2x²+16x+30

Use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=2(As, a>0 the parabola is open upward), b=16. by putting the values.

-b/2a = -16/2(2) = -4

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Now, find y- intercept put x=0 in the above equation. f(0)= 0+0+30, we get point (0, 30).

Now find x-intercept put y=0 in the above equation. 0=2x²+16x+30

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From vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. The graph is attached below.

5. y=(x+2)²+4

The general form of parabola is y=a(x-h)²+k , where vertex = (h,k)

if a>0 parabola is opened upward.

if a<0 parabola is opened downward.

Compare the given equation with general form of parabola.

-h=2 ⇒h=-2

k=4

so, vertex= (-2, 4)

As, a=1 which is greater than 0 so parabola is opened upward and the graph has minimum.

The graph is attached below.

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