Will give !
1. use the parabola tool to graph the quadratic function f(x)=x2+10x+16 . graph the parabola by first plotting its vertex and then plotting a second point on the parabola.

2. use the parabola tool to graph the quadratic function f(x)=−(x−3)(x+1) .
graph the parabola by first plotting its vertex and then plotting a second point on the parabola.

3. use the parabola tool to graph the quadratic function f(x)=−x2+4.
graph the parabola by first plotting its vertex and then plotting a second point on the parabola.

4. use the parabola tool to graph the quadratic function f(x)=2x2+16x+30 .
graph the parabola by first plotting its vertex and then plotting a second point on the parabola.

5. select all the statements that are true for the graph of y=(x+2)2+4 .

the graph has a maximum.

the graph has a minimum.

the vertex is (2, 4) .

the vertex is (−2, 4) .

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

f(-b/2a)= f(-4)= 2(-4)²+16(-4)+30 = -2

So, Vertex = (-4, -2)

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

2x²+16x+30=0 ⇒2(x²+8x+15)=0 ⇒x²+8x+15=0 ⇒x²+5x+3x+15=0 ⇒x(x+5)+3(x+5)=0 ⇒(x+5)(x+3)=0 ⇒x=-5 , x= -3

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.

the probability is 0

∠cbe and ∠deb are same-side interior angles.

i took the test