All categories

3 years ago

What is the vertex and x-intercepts of the graph of the function below y=x^2-6x-7

Mathematics

2 answers:

labwork [276]3 years ago

6 0

Answer: The vertex is (3, 16) and the x-intercepts are (-1, 0) and (7, 0).

Step-by-step explanation: We are given to find the vertex and x-intercepts of the graph of the following function :

$y=x^2-6x-7~~~~~~~~~~~~~~~~~~~~~~~~~~~`(i)$

We know that

the vertex of the graph of function $y=a f(x-h)^2-k$ is given by (h, k).

From equation (i), we have

$y=x^2-6x-7\\\\\Rightarrow y=(x^2-6x+9)-7-9\\\\\Rightarrow y=(x-3)^2-16.$

Therefore, the vertex is (3, 16).

The x-intercepts of function (i) will be given by

$x^2-6x-7=0\\\\\Rightarrow x^2-7x+x-7=0\\\\\Rightarrow x(x-7)+1(x-7)=0\\\\\Rightarrow (x+1)(x-7)=0\\\\\Rightarrow x+1=0,~~x-7=0\\\\\Rightarrow x=-1,~~x=7.$

Thus, the vertex is (3, 16) and the x-intercepts are (-1, 0) and (7, 0).

Strike441 [17]3 years ago

4 0

Answer:

x=(7,0),(-1,0) y=(0,-7)

Step-by-step explanation:

You might be interested in

Converting 52ft2 into m2

alexira [117]

Converting 52ft2 into meters sq

6 0

3 years ago

Damaris will be working at the local pool over his nine-week summer break. His net pay will be $123.56 each

lesya [120]

I wish I could help but I’m stúpid ;-;. Good luck finding the answer 00p

8 0

3 years ago

Write an equation that is parallel to y = 3x and passes through (-4 , -3)

Anit [1.1K]

The slopes of two parallel lines must be identical.

We have slope $m=3$ , so the slope for the parallel line be the same.

Now, to find an equation that also passes through the given point, we use slope-point form, $y-y_1=m(x-x_1)$ , where our point $(-4,-3)$ is substituted for $(x_1,y_1)$ .

$y+3=3(x+4)$

Now, we convert to slope-intercept form as such.

$y=3x+12-3\\y=3x+9$

And we are done. :) We can verify graphically that these are indeed parallel lines. See attached.

8 0

3 years ago

Can someone help me

erik [133]

Answer:

6 < x < 23.206

Step-by-step explanation:

To properly answer this question, we need to make the assumption that angle DAC is non-negative and that angle BCA is acute.

The maximum value of the angle DAC can be shown to occur when points B, C, and D are on a circle centered at A*. When that is the case, the sine of half of angle DAC is equal to 16/22 times the sine of half of angle BAC. That is, ...

(2x -12)/2 = arcsin(16/22×sin(24°))

x ≈ 23.206°

Of course, the minimum value of angle DAC is 0°, so the minimum value of x is ...

2x -12 = 0

x -6 = 0 . . . . . divide by 2

x = 6 . . . . . . . add 6

Then the range of values of x will be ...

6 < x < 23.206

_____

* One way to do this is to make use of the law of cosines:

22² = AB² + AC² -2·AB·AC·cos(48°)

16² = AD² + AC² -2·AD·AC·cos(2x-12)

The trick is to maximize x while satisfying the constraints that all of the lengths are positive. This will happen when AB=AC=AD, in which case the equations be come ...

22² = 2·AB²·(1-cos(48°))

16² = 2·AB²·(1 -cos(2x-12))

The value of AB drops out of the ratio of these equations, and the result for x is as above.

4 0

3 years ago

Read 2 more answers

The price of a swimming pool

Goryan [66]

Answer: