Ineed in this math problem, solve and explain to me because i have a test tommorrow. write the quadratic function in vertex form. then identify the vertex. y=5x²+10x+7

Vertex form: 5(x + 1)^2 + 2; Vertex: (-1, 2)

Step-by-step explanation:

To convert the quadratic function to vertex form you need to first isolate the variable x^2. To do this factor out 5 from only the first two terms in the equation.

y = 5(x^2 + 2x) + 7

Now use (b/2)^2 to complete the square, where b = 2.

Add 1 inside the parentheses and subtract 1 (times 5 since the 1 being added is also being multiplied by 5) outside the parentheses.

This equation:

5(x^2  + 2x) + 7

becomes:

5(x^2 + 2x + 1) + 7 - 1(5)

Factor the trinomial inside the parentheses by seeing what two of the same numbers add to b (2) and multiply to 1 (c). This number is a positive 1.

5(x + 1)^2 + 7 - 1(5)

Now simplify the entire equation.

5(x + 1)^2 + 2 is the vertex form of this quadratic function.

Vertex form is a(x - h)^2 + k, where (h, k) is the vertex. Since 1 and 2 are where h and k are in vertex form, these are your vertex coordinates.

However, the original form is a negative number, so this means that in order for it to be + 1 then the 1 has to be negative (two negative make a positive). Therefore;

The vertex is (-1, 2).

In quadratic function in vertex form:

y = 5x² + 10x + 7

= 5(x² + 2x) + 7

= 5[(x² + 2x + 1) - 1] + 7

= 5(x² + 2x + 1) - 1*5 + 7

= 5(x + 1)² - 5 + 7

= 5(x + 1)² + 2.

The vertex of a quadratic function in vertex form is:

(-1, 2)

The vertex form should look like this: q(x - p)² + b

So the vertex is (p, b). That is why the vertex of 5(x + 1)² + 2 is (-1, 2).

answer: 3.26019516

step-by-step explanation:

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)