Write the equation of the following graph in vertex form.

Mathematics

1 answer:

Marat540 [252]3 years ago

6 0

Answer:

Correct choice is 2. $f(x)=0.4(x+2)(x-5)$

Step-by-step explanation:

We have been given a graph of the quadratic function. Now we need to use that graph to find the equation of the graph in vertex form. By the way given choices written in intercept form so to be accurate we are going to find the equation in intercept form.

from graph we see that x-intercepts are at 5 and -2.

we know that if x=a is x-intercept then (x-a) must be factor.

So (x+2)(x-5) is the factor.

when leading coefficient is positive then graph opens upward.

so that means leading coefficient is 0.4

Hence correct choice is 2. $f(x)=0.4(x+2)(x-5)$

You might be interested in

Prove that: (a2 - b2)3 + (b2-c2)3+ (c2-a2)3 = 3 (a+b) (b+c) (c+a) (a-b) (b-c) (c-a).

Vanyuwa [196]

$L=(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3=(*)\\\\(a^2-b^2)^3=a^6-3a^4b^2+3a^2b^4-b^6\\\\(b^2-c^2)^3=b^6-3b^4c^2+3b^3c^4-c^6\\\\(c^2-a^2)^3=c^6-3c^4a^2+3c^2a^4-a^6\\\\(*)=a^6-3a^4b^2+3a^2b^4-b^6+b^6-3b^4c^2+3b^2c^4-c^6+c^6-\dots\\\dots-3c^4a^2+3c^2a^4-a^6\\\\=-3a^4b^2+3a^2b^4-3b^4c^2+3b^2c^4-3c^4a^2+3c^2a^4\\\\=3(-a^4b^2+a^2b^4-b^4c^2+b^2c^4-a^2c^4+a^4c^2)$

$R=3(a+b)(a-b)(b+c)(b-c)(c+a)(c-a)\\\\=3(a^2-b^2)(b^2-c^2)(c^2-a^2)\\\\=3(a^2b^2-a^2c^2-b^4+b^2c^2)(c^2-a^2)\\\\=3(a^2b^2c^2-a^4b^2-a^2c^4+a^4c^2-b^4c^2+a^2b^4+b^2c^4-a^2b^2c^2)\\\\=3(-a^4b^2+a^2b^4-b^4c^2+b^2c^4-a^2c^4+a^4c^2)\\\\L=R$

5 0

3 years ago

Read 2 more answers

The Pythagorean Theorem is

IrinaK [193]

False the Pythagorean theorem is A2+B2=C2 hope this helps!

5 0

3 years ago

Read 2 more answers

I need helppppp!!!!!!!

11111nata11111 [884]

Answer: