Use algebra to find the x and y intercepts , for a quadratic equation

**1. The problem statement, all variables and given/known data**

Use algebra to ﬁnd the x-intercepts and y-intercept of The graph of y=-x^2/5+x/5+6

this is a parabola.

**2. Relevant equations**

y=-x^2/5+x/5+6

**3. The attempt at a solution**

Find the y-intercept , which is substitute 0=x into the equation to find the answer , done that .

The problem I have is where to start to find the x-intercept , I know that y=0 so I now have x^2/5+x/5+6=0 , but believe I can't now factorise this equation and will have to first clear the fractions first . Is this the right place to start.

Re: Use algebra to find the x and y intercepts , for a quadratic equation

is it y=-x^2/5+x/5+6 or x^2/5+x/5+6=0?,

btw, for -x^2/5+x/5+6=0, then x^2-x-30=0, x = (1+/- 11)/2, ie x=6 or -5

Re: Use algebra to find the x and y intercepts , for a quadratic equation

Sorry missed the - sign in front of x^2 in my first reply. Method of course still applies.

Re: Use algebra to find the x and y intercepts , for a quadratic equation

The equation is y=-x^2/5+x/5+6 and I'm looking for the x-intercept .

Re: Use algebra to find the x and y intercepts , for a quadratic equation

Quote:

Originally Posted by

**duggielanger** The equation is y=-x^2/5+x/5+6 and I'm looking for the x-intercept .

Can you solve $\displaystyle x^2-x-30=0~?$ That is all that is needed.

Re: Use algebra to find the x and y intercepts , for a quadratic equation

I think I can , I would factorise this equation to get my final answer. which is (x-1)(x+5) I think, and these are my x-intercepts. But to get this part do I simply the fraction by finding the the products of 5 and 6 and then simplify to x^2-x-30=0.

And sorry one last question, would squaring and or using then quadratic formula get the same answer.

Re: Use algebra to find the x and y intercepts , for a quadratic equation

Quote:

Originally Posted by

**duggielanger** I think I can , I would factorise this equation to get my final answer. which is (x-1)(x+5) I think, and these are my x-intercepts. But to get this part do I simply the fraction by finding the the products of 5 and 6 and then simplify to x^2-x-30=0.

And sorry one last question, would squaring and or using then quadratic formula get the same answer.

It factors as $\displaystyle (x-6)(x+5)$. Thus your *x*-intercepts are $\displaystyle (6,0)~\&~(-5,0)($

Re: Use algebra to find the x and y intercepts , for a quadratic equation

Ok thank you , I will go back and study the section again to get to grips with it.