Question

Please help me identify the vertex and x-intercepts of f(x)=3/5(x^2+6x-5) and then check the results by writing the quadratic function in standard form??

Answer 1

@phi do you think you could help me with this one? its the same as the one we did earlier, but I can't find any way to factor the equation

Answer 2

that's because it does not factor (nicely)
are you sure you got the correct equation (x^2+6x PLUS 5 ) does factor)

otherwise we have to use the quadratic formula

Answer 3

no, sadly it is f(x)=3/5(x^2+6x-5)

Answer 4

then use the quadratic formula to find the x-intercepts. In other words, you solve for x when
\[ \frac{3}{5}\left(x^2+6x-5\right)=0 \]
we can multiply both sides by 5/3 to get rid of the 3/5
\[ \frac{5}{3}\cdot \frac{3}{5}\left(x^2+6x-5\right)=0 \cdot \frac{5}{3} \\
x^2+6x-5=0
\]

Answer 5

I got x^2+6x=5

x(x+6)=5

Answer 6

I think that it would be x=0 and x=-6 but im not sure what to do with the 5 on the other side

Answer 7

sadly, with a quadratic equation you can't solve it that way (as you just noticed)

Have you learned either "completing the square" or the quadratic formula ?

Answer 8

i've learned both ways but im not very good at them :o

Answer 9

let's try completing the square.
you got to
x^2+6x=5
now instead of factoring out an x, you "complete the square". find the number in front of the x term. divide that number by 2. square it. add the result to both sides of the equation.

Answer 10

but there is no nummber in front of the x term anymore right?

Answer 11

in other words, find the 6 in front of 6x. divide by 2: 6/2=3.
square 3 to get 3*3=9

Answer 12

oh i see

Answer 13

@ali1029