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I can't for the life of me figure out how to do this, some questions the method I use works and others it doesn't. One question that I am totally confused with is:

y = ${x}^2 + 2x -3 $

$for -2\leq x \leq 4$

Where you must state the range of the function over the given domain.

I'm the sort of student where if I am not given reasoning or steps/pattern/method to follow I won't be able to understand the concept correctly so if there is any pattern or method you guys use to solve these sort of questions I'd love to hear it.

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    The function $f$ is a quadratic function. Have you studied quadratic functions? If not, then you're missing prerquisite knowledge for this problem. If you _have_ studied quadratic functions, what can you say about the graph of $y=f(x)$? – 2017-02-19
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    Factoring $f(x) = x^2+2x-3 = (x-1)(x+3)$ to find the roots $-1,3$ of $f$ might help. You should also know what quadratic functions look like. – 2017-02-19
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    Partially, personally I'd let y = 0 then factorize such that 0 = (x+3)(x-1), and the two x intercepts of the parabola are x = -3 and 1 but from then I am lost. – 2017-02-19
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    Do you know how to derive? And what the result of the derivative can tell about the function? – 2017-02-19
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    So far only due to my external studies do I know how to use first principles to get the derivative of a function but I've never used it for these sort of questions, I was only using the first derivative for finding the point that a tangent touches a curve and the angle between two tangents etc. – 2017-02-19
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    Equation of a tangent to curve is what I meant not the coordinates that a tangent touches a curve – 2017-02-19
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    Knowing that it's a quadratic function, and that its main coefficient is positive, why don't you look for its vertex and if it is between -2 and 4 you know which is the lowest value that it can take. – 2017-02-19
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    Apparently I am stupid and was looking at the incorrect answer, smh... I had the correct answer the entire time. $-3\leq y \leq 21$ – 2017-02-19

2 Answers 2

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In your algebra / precalculus course you might be using methods such as "completing the square" and "vertex form" to study properties of quadratics.

Let's complete the square on the quadratic $f(x) = x^2 + 2x -3$ so we can put $f(x)$ into vertex form.

$f(x) + 3 = x^2 + 2x$

$f(x) + 3 + 1 = x^2 + 2x + 1$

$f(x) + 4 = (x+1)^2$

$f(x) = (x+1)^2 - 4.$

If this vertex form is something you are familiar with, then we know we must have a vertex on our quadratic at $(-1,-4)$, and $-1$ is in the domain we are interested in!

But, since the quadratic we originally started with, $f(x) = x^2 + 2x - 3$ is a parabola opening upwards, we know that the vertex is going to be the lowest point in the range on the domain we are considering. Now we just need to find the highest point, so we evaluate the endpoints of the domain we are considering. We check manually:

$f(-2) = (-2+1)^2 - 4 = -3$

$f(4) = (4+1)^2 - 4 = 5^2 -4 = 25 - 4 = 21.$

Now we have our highest point in the range, 21.

Since polynomials are continuous, we know we will hit every point from -4, up to 21, giving the range as [-4,21].

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It should be [-4,21].

Derivative of the function is 2x+2
Let it be equal to zero
2x+2=0
x=-1
this means function has maxima or minima at x=-1
Now find values of function at -2,4 and -1, which are -3,21 and -4 resp.
So the range is -4<=y<=21

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    The OP tagged this as pre-calculus. They probably do not have derivatives at their disposal. You should probably find the vertex using elementary methods and use alternative analytic geometric reasoning. – 2017-02-19