Question 130393
First let's find the y-intercept. To do that, simply plug in {{{x=0}}}




{{{f(x)=5x^2-38x+21}}} Start with the given function



{{{f(0)=5(0)^2-38(0)+21}}} Plug in {{{x=0}}}



{{{f(0)=5*0-38*0+21}}} Raise 0 to the 2nd power to get 0



{{{f(0)=0-38*0+21}}} Multiply 5 and 0 to get 0



{{{f(0)=0-0+21}}} Multiply 38 and 0 to get 0



{{{f(0)=0+21}}} Subtract 0 from 0 to get 0



{{{f(0)=21}}} Add 0 and 21 to get 21




So the y-intercept is (0,21)



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Now let's find the x-intercept. To do that, simply plug in {{{f(x)=0}}} (ie {{{y=0}}})



{{{f(x)=5x^2-38x+21}}} Start with the given function



{{{0=5x^2-38x+21}}} Plug in {{{f(x)=0}}}





{{{0=(x-7)(5x-3)}}} Factor the right side (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)




Now set each factor equal to zero:

{{{x-7=0}}} or  {{{5x-3=0}}} 


{{{x=7}}} or  {{{x=3/5}}}    Now solve for x in each case



So the x-intercepts are 

*[Tex \LARGE \left(7,0\right)] and  *[Tex \LARGE \left(\frac{3}{5},0\right)]