# Math Help - HELP! Quadratic Equations.

How do you find the zeros and the lines of symmetry for the graphs of the following functions:

1. $f(x) = x^2 - 4x + 1$
2. $g(x) = x^2 + 6x - 11$
3. $h(x) = x^2 - 24$

I know how to get the zeros for 1 and 2, but not 3. Is for #1 x = 3.87, 0.31 and for #2 x=1.47, 7.47 ?
I don't get what the line of symmetry is at all...

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How do you find the minimum and maximum value of these functions:

• $f(x) = x^2 - 3x + 9$
• $g(x) = -x^2 + 8x +2$
• $h(x) = x^2 - 49$

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Thank you!

2. Originally Posted by ninjuhtime
How do you find the zeros and the lines of symmetry for the graphs of the following functions:

1. $f(x) = x^2 - 4x + 1$
2. $g(x) = x^2 + 6x - 11$
3. $h(x) = x^2 - 24$
For number 3, you can either use the quadratic formula taking a=1, b=0, c=-24 or you can set the equation equal to zero and solve:

$x^2-24=0$
$x^2=24$

Then take the square root of each side and simplify the square root on the right.

I know how to get the zeros for 1 and 2, but not 3. Is for #1 x = 3.87, 0.31 and for #2 x=1.47, 7.47 ?

I don't get what the line of symmetry is at all...

---------------------------------

How do you find the minimum and maximum value of these functions:

• $f(x) = x^2 - 3x + 9$
• $g(x) = -x^2 + 8x +2$
• $h(x) = x^2 - 49$

You can to it one of two ways... If you are in calculus, take the derivative and find the critical values.

If you are not in calculus, think about what that graph looks like...it's a parabola right? Parabolas always have a maximum or minimum...and it occurs at the....vertex. The x-coordinate of the vertex is given by $-b/(2a)$. So once you find the x-coordinate, you can put that x value back into the function to find the y-value at that point. To think about if it's a minimum or maximum you need to figure out if the parabola is opening upwards or downwards....hint: think about the sign of the leading coeffecient, the sign on the $x^2$ term.

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Thank you!
Hope this helps!