Use the quadratic formula to find the x-intercepts:
y=-(x-10)^2+140

Question

Use the quadratic formula to find the x-intercepts:
y=-(x-10)^2+140

welshfella · Answer

first  rearrange the equaton to the form ax^2 + bx + c = 0

mathstudent55 · Answer

The x-intercepts occur where y = 0.

mathstudent55 · Answer

First, replace y with 0.
Then write the quadratic in standard form, as @welshfella showed above.
Then you can read the coefficients a, b, and c, and use the quadratic formula.

sparklyme · Answer

@welshfella @mathstudent55 So would it be$$-x ^{2}-100+140=0$$

welshfella · Answer

-(x - 10)^2 + 140
= -  ( x^2 - 20x + 100) + 140
= -x^2 + 20x - 100 + 140
= -x^2  + 20x + 40

sparklyme · Answer

@welshfella @mathstudent55 where did you get the 20x?

mathstudent55 · Answer

Here it is step by step:

$y=-(x-10)^2+140$

$0 = -(x-10)^2+140$

$-(x-10)^2+140 =0$

$-(x^2 - 20x + 100) + 140 = 0$

$-x^2 + 20x - 100 + 140 = 0$

$-x^2 + 20x + 40 = 0$

$x^2 - 20x - 40 = 0$

sparklyme · Answer

Where did you get 20x?

mathstudent55 · Answer

Notice how I dealt with $(x - 10)^2$

It is the square of a binomial and follows the pattern:

$(a - b)^2 = a^2 - 2ab + b^2$

mathstudent55 · Answer

If you don't remember that, you can always go back to

$(a - b)^2 = (a - b)(a - b)$

and use FOIL

sparklyme · Answer

Yea. I like that better thanks.

sparklyme · Answer

So what would I FOIL? The whole thing or just what is inside the ()?

mathstudent55 · Answer

You just use a product of binomials to handle the square of the binomial.

$(x - 10)(x - 10) = x^2 -10x - 10x + 100 = x^2 - 20x + 100$

sparklyme · Answer

ohhh....right. thank you so much for everything. it means a lot!

mathstudent55 · Answer

I'll show you the steps again using the product of the binomials instead of the square of the binomial.

sparklyme · Answer

Okay!

mathstudent55 · Answer

We start out just as we did before.

$y=-(x-10)^2+140$

Replace y with 0.

$0 = -(x-10)^2+140$

Switch sides to have zero on the right side.

$-(x-10)^2+140 = 0$

Now show the product of binomials instead of the square of the binomial.

$- (x - 10)(x - 10) +140 = 0$

mathstudent55 · Answer

Now we use FOIL.
Remember we must keep that negative sign to the left of the the product of the binomials.
We will deal with the negative sign after we use FOIL.

Here is the last line from above again.

$- \color{red}{(x - 10)(x - 10)} +140 = 0$

Now we use FOIL keeping the negative sign to the left as it must be kept.

$- \color{red}{(x^2 - 10x -10x + 100)} +140 = 0$

Now we combine like terms after FOIL.

$- \color{red}{(x^2 - 20x + 100)} +140 = 0$

Now we deal with the negative outside the parentheses.
To distribute a negative sign, think of the negative sign as being -1 multiplying the parentheses.
We multiply everything inside the parentheses by -1.
What that does is change every sign inside the parentheses.

$- x^2 + 20x - 100 +140 = 0$

Ok so far?

sparklyme · Answer

Yes.

sparklyme · Answer

After this I think I know what to do. Thank you so much for everything!

mathstudent55 · Answer

Now we combine like terms, -100 + 140 = 40

$- x^2 + 20x +40 = 0$

Now I multiply everything on both sides by -1 to get rid of the negative sign on the x^2 term.

$x^2 - 20x - 40 = 0$

This last step is just my preference.
You can use the previous line with first term -x^2 already in the quadratic formula.

Now you are ready for the quadratic formula.

mathstudent55 · Answer

You're welcome.