Question

4) Write the quadratic equation of the following in vertex form: vertex (-1,3) and passes through (1,-5)

I
ULTS
Factor completely
5) 15m2 - 6
6) 2x2 + 5x - 10% - 25

Answer 1

You will have to set up a system of equations

Answer 2

Please check the explanation.

Explanation:

4)

If an equation representing a parabola is in vertex form such as

`y\:=a\left(x-k\right)^2+h`

then its vertex will be at (k, h).

Therefore the equation for a parabola with a vertex at (-1, 3), will have the general form

`y\:=a\left(x+1\right)^2+3`

If this parabola also passes through the point (1, -5) then we can determine the 'a ' parameter.

`-5\:=a\left(1+1\right)^2+3`

simplifying the equation

`2^2a+3=-5`

`4a+3=-5`

subtract 3 from both sides

`4a+3-3=-5-3`

`4a=-8`

Divide both sides by 4

`(4a)/(4)=(-8)/(4)`

`a=-2`

So our equation in vertex form is:

`y\:=-2\left(x+1\right)^2+3`

5)

Given the expression

`15n^2-6n`

`\mathrm{Apply\:exponent\:rule}:\quad \:a^(b+c)=a^ba^c`

`=15nn-6n`

`\mathrm{Rewrite\:}6\mathrm{\:as\:}3\cdot \:2`

`\mathrm{Rewrite\:}15\mathrm{\:as\:}3\cdot \:5`

`=3\cdot \:5nn-3\cdot \:2n`

Factor out the common term 3n

`=3n\left(5n-2\right)`

6)

Given the expression

`2x^2+5x-10x-25`

Factor 2x²+5x: x(2x+5)

Factor -10x-25: -5(2x+5)

so the expression becomes

`=x\left(2x+5\right)-5\left(2x+5\right)`

`\mathrm{Factor\:out\:common\:term\:}\left(5+2x\right)`