Link between Quadratic function and gradient at a specific point on the X axis?

Question

Take any quadratic function, and graph it, for example y=x^2, y=2x^2 or y=2x^+4x, then at a specific point on the X axis, draw a tangent to the graph, how can you work out the gradient of the function?

Answer 1

I don't know if you learned it yet, but the derivative (calculus) gives you the slope of the tangent at any value of x on the curve. The power rule is that x^n = nx^(n-1), and the derivative of a sum is the sum of the derivatives of the components. So for example, if y=x^2, then dy/dx = 2x. Then the slope at point x is equal to 2x. Similarly, for y=2x^2, dy/dx=4x, and for y=2x^2+4x, dy/dx = 4x+4.

Answer 2

If the parabola has real roots at x1 and x2, then no tangent can be drawn to the parabola from any point on the x-axis that lies between x1 and x2.

The problem is that you can't accurately draw a tangent from some point (x3,0) to the parabola with any degree of accuracy. You must know the point on the parabola to which you wish to draw a tangent and that will determine where the tangent will cross the x-axis.

The slope of the parabola at any point on the parabola is given by slope = 2ax +b.

If your equation were y=x^2+2x-15, then at the point(1,-12) on the parabola, the slope would be 2*1*1+2) = 4. The equation of the tangent would be y = 4x + b. Find b by putting y=-12 and x =1 in the above equation getting b= -16

So the equation would be y = 4x-16. This line woul have an x- intercept of x = (4,0). Notice it is not in the range -5
If the parabola has imaginary roots, then there is no restriction on x.

Answer 3

If the tangent is drawn from the point (ξ, 0) then its equation will be y = m(x - ξ), where m is the gradient
If the equation of the quadratic is y = ax² + bx +c, then, where the tangent touches,
m(x - ξ) = ax² + bx +c
which rearranges to:
ax² + (b - m)x + c + mξ = 0
For the tangent to touch rather than cross, this equation must have two identical solutions => the discriminant must equal zero.
So (b - m)² - 4a(c + mξ) = 0
=> m² - 2(b + 2ξ)m + b² - 4ac = 0
which, in general, will solve to give two values of m.

As someone else pointed out, it will not always be possible to draw tangents. in which case the solutions for m will not be real.

Answer 4

The easiest and most accurate way is to use differentiation.
if you differentiate y = x^2, you get dy/dx = 2x
and this latter function gives you the gradient at any value of x.
So for y=x^2, gradient = 2x. Just plug in whatever value of x you want.

for y=2x^2, gradient = 4x.
y=2x^2+4x, gradient = 4x + 4

Answer 5

It is simply a vector perpendicular to the tangent
take for example y=x^2 then function of x and y
f(x,y)=y-x^2=0
its components are deivative of f with respect to x given y is constant and derivative of f with respect to y given x constant so (-2x,1) so at a point with coordinates (x0,y0) the gradient is (-2x0,1)
See also the external links in
http://en.wikipedia.org/wiki/Gradient

Answer 6

Q1: f(x)=2x^2-4x+7 =2(x^2-2x+a million)-2+7 =2(x-a million)^2+5 Q2: the minimum element is obtrusive (a million,5), you may get to are conscious of it by utilising drawing the chart or the two by utilising right this moment viewing the consequence of Q1. wish it effective for it...gud success

Answer 7

the gradient of a straight line y = mx +b is constant ie m
since a quadratic is a curve its gradient is changing
the gradient of a curve is it's derivative
so take y = x^2
then the derivative ( or gradient ) is
m = 2x so the gradient depends on x aswell
so if x = 1,2,3 ....etc
m= 2,4,6 .......etc
so the graph is continuously getting steeper

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Answer 8

The answer to this question is to locate two points on the tangent drawn and divide the vertical difference of these two points by their horizontal difference using their coordinates. This is the easiest and by far the most straight-forward way, instead of using their derivatives.