Answer:
[tex]x[/tex]-intercepts at [tex](1,0)[/tex] and [tex](0.8,0)[/tex]
Step-by-step explanation:
A quadratic in the form [tex]ax^2+bx+c=y[/tex] crosses the [tex]x[/tex]-axis when [tex]y=0[/tex].
The first step is replacing [tex]f(x)[/tex] with [tex]y[/tex].
So: [tex]y= -5x^2+9x-4[/tex]. With the information above, we can find the [tex]x[/tex] intercepts by setting [tex]y = 0[/tex].
Therefore [tex]0=-5x^2+9x-4[/tex].
Now we can use the quadratic formula because it is in the form [tex]ax^2+bx+c=0[/tex].
Note the quadratic formula: [tex]\frac{-b\frac{+}{-}\sqrt{b^2-4ac} }{2a } = x[/tex].
To find the values of [tex]a[/tex],[tex]b[/tex] and [tex]c[/tex] we can compare the equation to the general equation.
Therefore: [tex]a=-5[/tex], [tex]b=9[/tex] and [tex]c=-4[/tex].
Now put these values into the quadratic formula:
[tex]\frac{-9\frac{+}{-}\sqrt{(9)^2-4(-5)(-4)} }{2(-5) } = x[/tex]
And simplify:
[tex]\frac{-9\frac{+}{-}\sqrt{81-80} }{-10 }[/tex] , [tex]\frac{-9\frac{+}{-}\sqrt{1} }{-10 }[/tex].
[tex]\sqrt{1} =1[/tex], therefore [tex]x = \frac{-9\frac{+}{-}1 }{-10 }[/tex]
Now we can have two values for [tex]x[/tex]. One when we take away the discriminant ([tex]b^2-4ac[/tex]) and one when we add it.
So [tex]x = \frac{-10}{-10} = 1[/tex]
or
[tex]x = \frac{-8}{-10} = 0.8[/tex]
Therefore [tex]x[/tex]-intercepts at [tex](1,0)[/tex] and [tex](0.8,0)[/tex]
