What are the x- and y-interepts of the graph of 5x-4y=20?

Mathematics

1 answer:

77julia77 [94]3 years ago

8 0

Answer: x=4,y=-5

Step-by-step explanation:

Equation: 5x-4y=20

To find the x intercept make y=0 so ur left with

1)5x-4(0)=20

Then you can solve x from here

2) 5x=20

Because -4 * 0 = 0

3) x= 20/5

Because we want x by itself so we divide 5 on both side so the 5 cancels out on the left side

4) x=4

Because 20/5=4

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You do the same thing for y make x equal to zero and that’s how you find it

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Y=x^2-6x-16 in vertex form

satela [25.4K]

Answer:

$y=(x-3)^{2} -25$

Step-by-step explanation:

The standard form of a quadratic equation is $y=ax^{2} +bx+c$

The vertex form of a quadratic equation is $y=a(x-h)^{2} +k$

The vertex of a quadratic is (h,k) which is the maximum or minimum of a quadratic equation. To find the vertex of a quadratic, you can either graph the function and find the vertex, or you can find it algebraically.

To find the h-value of the vertex, you use the following equation:

$h=\frac{-b}{2a}$

In this case, our quadratic equation is $y=x^{2} -6x-16$ . Our a-value is 1, our b-value is -6, and our c-value is -16. We will only be using the a and b values. To find the h-value, we will plug in these values into the equation shown below.

$h=\frac{-b}{2a}$ ⇒ $h=\frac{-(-6)}{2(1)}=\frac{6}{2} =3$

Now, that we found our h-value, we need to find our k-value. To find the k-value, you plug in the h-value we found into the given quadratic equation which in this case is $y=x^{2} -6x-16$

$y=x^{2} -6x-16$ ⇒ $y=(3)^{2} -6(3)-16$ ⇒ $y=9-18-16$ ⇒ $y=-25$