# Finding line of symmetry and x-intercepts in simple quadratic

Ok so in a non calculator (mental) practice question. I've been given a quadratic equation in the form $$y = ax^2 + bx + c$$:

$$y= 2 + 1.75x - 0.25x^2$$

After rearranging and changing the decimals to fractions I end up with:

$$y = (-1/4)x^2 + (7/4)x + 2$$

I'm then asked to determine the equation of the line of symmetry (LOS) and then the coordinates of the x-intercepts. To work out the LOS, I used the formula $$-b/(2a)$$ and my answer was $$x= 28/8$$ ($$3.5$$). However I don't think this answer is correct and I also don't know how to get the x-intercepts from this.

If anyone could explain (in simple steps) how to get the right answers it would be helpful.

I graphed it on my calculator (even though i'm not supposed to) and it looks like the x-intercepts are near $$-1$$ and $$8$$ according to Wolfram Alpha.

• Don't get confused between the line of symmetry (the average of the roots), the roots (where the graph meets the x-axis, or when these values produce 0 when plugged into the equation), and the discriminant (a value that shows the number of real roots of the equation)! – Toby Mak May 27 '17 at 8:22
• Your axis of symmetry is correct. Why did you doubt it? – amd May 27 '17 at 18:47

It seems like you have confused the line of symmetry with the roots of the quadratic. The line of symmetry expresses the average of the two roots (that can be real or complex) since a parabola is symmetrical in the x-direction. When you average the two roots and get a value $a$, the line of symmetry is at $x = a$.

To find the real roots, you need to find the sum of roots $-b/a$, and the product of roots $c/a$ to get the roots $x_1$ and $x_2$. In this case, $x_1 + x_2 = 7$, and $x_1 x_2 = -8$. If we subtract both sides of the first equation by $x_2$, we get $x_1 = 7 - x_2$. Substituting the value into the second equation we have:

$$(7-x_2)(x_2) = 8$$ $$-{x_2}^2 + 7{x_2} - 8 = 0$$ $${x_2}^2 - 7{x_2} + 8 = 0$$ $$(x_2 + 1)(x_2 - 8) = 0$$ $$x_2 = -1,8$$

Solving for $x_1$ we have $x_1 = 7 - (-1) = 8$, or $x_1 = 7 - (+8) = -1$. However in this case, the other values of $x_1$ and $x_2$ are exactly the same ones as before, except they are swapped! So the only solutions are $x = -1,8$.

Hope this non-standard method helped. This method works for all quadratics, so have a go using this method when you see another problem like this!

• A point to note: If the line of symmetry were used as the real root, then every parabola would have only one real root, an obvious contradiction. – Toby Mak May 27 '17 at 8:27

Another approach: as $-\frac{1}{4}x^2+\frac{7}{4}x=0\iff x=0$ or $x=7$ we observe that $f(0)=f(7)=2$, hence the first coordinate of the vertex is $7/2$. As $f(7/2)=81/16$, we have $$f(x)=-\frac{1}{4}(x-7/2)^2+\frac{81}{16}.$$ From here the zeros are easily calculated.