Can someone show me the steps of working out:

The X Axis intercepts

If the function of the graph is given.

Or any other websites where I could learn these.

Thanks,

Dru

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- Feb 23rd 2009, 09:37 PM22upon7[SOLVED] Quadratics/Parabolas Help
Can someone show me the steps of working out:

The X Axis intercepts

If the function of the graph is given.

Or any other websites where I could learn these.

Thanks,

Dru - Feb 23rd 2009, 10:07 PMGrandadQuadratics/Parabola
Hello 22upon7I'm assuming that the equation you're referring to is in the form $\displaystyle y = ax^2 +bx + c$. As an example, I'll use $\displaystyle y = 2x^2 - 7x +3$. So:

(1) The coordinates of the turning point are found as follows:

- Differentiate, and find the value of $\displaystyle x$ for which $\displaystyle \frac{dy}{dx} =0$
- Substitute this value of $\displaystyle x$ in the original equation to find the value of $\displaystyle y$

So: $\displaystyle y = 2x^2 - 7x +3$

$\displaystyle \Rightarrow \frac{dy}{dx}=4x -7 =0$ when

$\displaystyle x = \frac{7}{4}$ and $\displaystyle y = 2\times \frac{49}{16}-7\times \frac{7}{4}+3 = -\frac{25}{8}$

So the turning point is $\displaystyle \left(\frac{7}{4}, -\frac{25}{8}\right)$, and it is a minimum since $\displaystyle \frac{d^2y}{dx^2}=4>0$.

(2) The equation of the line of symmetry is simply the line through the turning point parallel to the $\displaystyle y$-axis.

In this example, then, this is $\displaystyle x = \frac{7}{4}$.

(3) The $\displaystyle x$-axis intercepts are the values of $\displaystyle x$ for which $\displaystyle y = 0$

So: $\displaystyle y =0$ when $\displaystyle 2x^2 -7x +3 = 0$

$\displaystyle \Rightarrow (2x-1)(x-3)=0$

$\displaystyle \Rightarrow x = \tfrac{1}{2}$ or $\displaystyle 3$

So the intercepts are $\displaystyle (\tfrac{1}{2}, 0)$ and $\displaystyle (3,0)$.

Grandad - Feb 23rd 2009, 10:20 PM22upon7
- Feb 24th 2009, 12:02 PMGrandadDifferentiating
Hello 22upon7Have you covered differentiation in your course yet? This is an very important topic with many and varied applications, the simplest of which is to find a formula for the gradient of a curve at a particular point.

You'll probably need a teacher to explain it to you, and this forum isn't the place to detail all that you need to know. However, here's a web-site with a summary that you might find helpful: Differentiation

Grandad - Feb 24th 2009, 01:15 PMmasters
Hi 22upon7,

Grandad showed you how to factor and find the x-intercepts.

Let me show you another way to find the vertex (turning point) without differentiation. Your posting history indicates you would like to see an algebraic solution.

If you need to find the vertex of this parabola, first find the x-coordinate of the vertex by adding the x-intercepts and dividing the result by 2. This is the same thing as finding the midpoint between the two x-intercepts. Parabolas are symmetrical, so the axis of symmetry would run through this point.

There's also a generic formula for finding the x-coordinate of the vertex. It's $\displaystyle x=\frac{-b}{2a}$ from the general form $\displaystyle ax^2+bx+c=0$

So using either method, we find the x-coordinate of the vertex to be $\displaystyle \frac{7}{4}$.

To find the y-coordinate of the vertex, find $\displaystyle f\left(\frac{7}{4}\right)=2x^2-7x+3$

$\displaystyle f\left(\frac{7}{4}\right)=-\frac{25}{8}$

So the vertex of this parabola is $\displaystyle \left(\frac{7}{4}, -\frac{25}{8}\right)$