Question

Please help!!!

Solve (find x intercepts) using the quadratic formula

f(x)= -5x^+9x-4​ ​

Answer 1

`x`-intercepts at
`(1,0)` and
`(0.8,0)`

Explanation:

A quadratic in the form
`ax^2+bx+c=y` crosses the
`x`-axis when
`y=0`.

The first step is replacing
`f(x)` with
`y`.

So:
`y= -5x^2+9x-4`. With the information above, we can find the
`x` intercepts by setting
`y = 0`.

Therefore
`0=-5x^2+9x-4`.

Now we can use the quadratic formula because it is in the form
`ax^2+bx+c=0`.

Note the quadratic formula:
`(-b(+)/(-)√(b^2-4ac) )/(2a ) = x`.

To find the values of
`a`,
`b` and
`c` we can compare the equation to the general equation.

Therefore:
`a=-5`,
`b=9` and
`c=-4`.

Now put these values into the quadratic formula:

`(-9(+)/(-)√((9)^2-4(-5)(-4)) )/(2(-5) ) = x`

And simplify:

`(-9(+)/(-)√(81-80) )/(-10 )` ,
`(-9(+)/(-)√(1) )/(-10 )`.

`√(1) =1`, therefore
`x = (-9(+)/(-)1 )/(-10 )`

Now we can have two values for
`x`. One when we take away the discriminant (
`b^2-4ac`) and one when we add it.

So
`x = (-10)/(-10) = 1`

or

`x = (-8)/(-10) = 0.8`

Therefore
`x`-intercepts at
`(1,0)` and
`(0.8,0)`