Quadratic Formula
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blamocur
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If I were grading your work I'd have two issues:
- First you say [imath]\Delta = b^2 - 4ac[/imath], then [imath]\Delta = b^2+4ac[/imath], which makes no sense to me.
- You are right that [imath]\Delta > 0[/imath], but you can add to the description of the graph by comparing values of [imath]y[/imath] for [imath]x=0[/imath] and for large values of [imath]x[/imath].
Two things can always be done with [imath] y=ax^2+bx+c [/imath]. One is the completion of the square:
[math]\begin{array}{lll} y&=ax^2+bx+c=a\left(x^2+\dfrac{b}{a}x+\dfrac{c}{a}\right)=a\left(x+\dfrac{b}{2a}\right)^2-\dfrac{b^2-4ac}{4a} \end{array}[/math]which already gives you the information on whether the parabola is open at the top or bottom and whether it has zeroes.
It is also why we consider [imath] \Delta = b^2 -4ac[/imath]. It decides whether [imath] y=0 [/imath] has none, one or two solutions.
The other is differentiation. [imath] y'(x)=2ax+b [/imath] and [imath] y''(x)=2a. [/imath] The solution of the equation [imath] y'(x)=0 \Longleftrightarrow x=-\dfrac{b}{2a}[/imath] tells us where the parabola has its maximum or minimum, depending on whether [imath] y''(x)=2a [/imath] is positive or negative. In your case, it is assumed to be negative, so the parabola is open at the bottom.
Mnemonic: [imath]y=x^2 [/imath] opens at the top, so all negative parabolas [imath] y=-x^2+\ldots [/imath] open at the bottom.
[math]\begin{array}{lll} y&=ax^2+bx+c=a\left(x^2+\dfrac{b}{a}x+\dfrac{c}{a}\right)=a\left(x+\dfrac{b}{2a}\right)^2-\dfrac{b^2-4ac}{4a} \end{array}[/math]which already gives you the information on whether the parabola is open at the top or bottom and whether it has zeroes.
It is also why we consider [imath] \Delta = b^2 -4ac[/imath]. It decides whether [imath] y=0 [/imath] has none, one or two solutions.
The other is differentiation. [imath] y'(x)=2ax+b [/imath] and [imath] y''(x)=2a. [/imath] The solution of the equation [imath] y'(x)=0 \Longleftrightarrow x=-\dfrac{b}{2a}[/imath] tells us where the parabola has its maximum or minimum, depending on whether [imath] y''(x)=2a [/imath] is positive or negative. In your case, it is assumed to be negative, so the parabola is open at the bottom.
Mnemonic: [imath]y=x^2 [/imath] opens at the top, so all negative parabolas [imath] y=-x^2+\ldots [/imath] open at the bottom.
Two theorems come to my mind regarding this topic that don't invoke calculus.
120. For all real values of [imath]x[/imath], the expression [imath]ax^2+bx+c[/imath] has the same sign as [imath]a[/imath], except when the roots of the equation [imath]ax^2+bx+c=0[/imath] are real and unequal, and [imath]x[/imath] has a value lying between them.
121. From the preceding article, it follows that the expression [imath]ax^2+bx+c[/imath] will always have the same sign whatever real value x may have, provided [imath]b^2-4ac[/imath] is negative or zero; and if this condition is satisfied the expression is positive or negative according as [imath]a[/imath] is positive or negative. Conversely, so that the expression [imath]ax^2+bx+c[/imath] may be always positive, [imath]b^2-4ac[/imath] must be negative or zero, and [imath]a[/imath] must be positive; and in order that [imath]ax^2+bx+c[/imath]may be always negative [imath]b^2-4ac[/imath] must be negative or zero, and [imath]a[/imath] must be negative.
120. For all real values of [imath]x[/imath], the expression [imath]ax^2+bx+c[/imath] has the same sign as [imath]a[/imath], except when the roots of the equation [imath]ax^2+bx+c=0[/imath] are real and unequal, and [imath]x[/imath] has a value lying between them.
121. From the preceding article, it follows that the expression [imath]ax^2+bx+c[/imath] will always have the same sign whatever real value x may have, provided [imath]b^2-4ac[/imath] is negative or zero; and if this condition is satisfied the expression is positive or negative according as [imath]a[/imath] is positive or negative. Conversely, so that the expression [imath]ax^2+bx+c[/imath] may be always positive, [imath]b^2-4ac[/imath] must be negative or zero, and [imath]a[/imath] must be positive; and in order that [imath]ax^2+bx+c[/imath]may be always negative [imath]b^2-4ac[/imath] must be negative or zero, and [imath]a[/imath] must be negative.
p.90 Higher Algebra by Hall & Knight.
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It is common to refer to the graph of a quadratic function in the case of [imath]a>0[/imath] as an up parabola, and in the case of [imath]a<0[/imath] as a down parabola. However, that assumes that the definition of a parabola is known in the first place. First, one must prove that the graph of a quadratic function is a parabola.
Harry_the_cat
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Yes you can show it algebraically as you have attempted. Steven G has shown you how to fix it up, so each statement is correct.
Alternatively, you could think about it graphically:
If a<0, then the graph is an "upside-down" parabola.
If c>0, then, seeing that c is the y-intercept, the parabola cuts the y-axis above the origin.
Now, stop and think about those 2 things. Can you see that the graph MUST have 2 x-intercepts? (Try to draw it otherwise.)
Note also: You concluded "there are 2 solutions to the graph in the question".
Graphs don't have solutions. Equations do though.
You should have concluded:
There are 2 solutions to the equation \(\displaystyle ax^2+bx+c=0\) which means that the graph of \(\displaystyle y=ax^2+bx+c\) has 2 x-intercepts.
Alternatively, you could think about it graphically:
If a<0, then the graph is an "upside-down" parabola.
If c>0, then, seeing that c is the y-intercept, the parabola cuts the y-axis above the origin.
Now, stop and think about those 2 things. Can you see that the graph MUST have 2 x-intercepts? (Try to draw it otherwise.)
Note also: You concluded "there are 2 solutions to the graph in the question".
Graphs don't have solutions. Equations do though.
You should have concluded:
There are 2 solutions to the equation \(\displaystyle ax^2+bx+c=0\) which means that the graph of \(\displaystyle y=ax^2+bx+c\) has 2 x-intercepts.