1. ## graphs

Hello, I am really stuck on these 2 questions could someone help please, much appreciated !

1)

2) (I have attemped this one but I do not know if it is right!)

Need Help

2. In your parabola problem, set y = 0 and solve the quadratic equation. These will be the x-intercepts.

$-2x^2+5x+7=0$

$2x^2-5x-7=0$

Can you finish?

In your second rational equation, you made a mistake at the end of what is shown there. You got a sign wrong.

$x^2-x-2x{\color{red}-2}=x^2-1$

Can you finish from here?

3. Originally Posted by masters
In your parabola problem, set y = 0 and solve the quadratic equation. These will be the x-intercepts.

$-2x^2+5x+7=0$

$2x^2-5x-7=0$

Can you finish?

In your second rational equation, you made a mistake at the end of what is shown there. You got a sign wrong.

$x^2-x-2x{\color{red}-2}=x^2-1$

Can you finish from here?
Thanks but what do I do after both questions...

1) i.e how do I solve the equation to get the x coordinates?

2) I'm completely stuck! (I donot know what to do after my attempt)

4. Originally Posted by need help
Thanks but what do I do after both questions...

1) i.e how do I solve the equation to get the x coordinates?

2) I'm completely stuck! (I donot know what to do after my attempt)

To find the x-coordinates in your first problem, you must solve the quadratic equation for its two values for x. You can do this by factoring, if possible, or by using the quadratic formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

But $2x^2-5x-7=0$ factors quite nicely into $(2x-7)(x+1)=0$,

Set each of those factors equal to zero and we have our intercepts.

$2x-7=0 \ \ or \ \ x+1=0$

$\boxed{x=\frac{7}{2}} \ \ or \ \ \boxed{ x=-1}$

I suggest you check each of these answers by substituting them back into the original quadratic equation.

To finish your second rational equation from where you left it,

$x^2-x-2x-2=x^2-1$

The $x^2$ cancels out and we have

$-3x=1$

$\boxed{x=\frac{1}{3}}$

Check this solution by substituting back into the original rational equation.