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(a) Draw the graph of the parabola. (b) From your graph, estimate the $x$ -intercepts. (c) Check your estimate by finding the $x$ -intercepts exactly.$$y=3(x+1)^{2}+4$$

(b) None(c) None

Algebra

Chapter 1

Functions and their Applications

Section 4

Quadratic Functions - Parabolas

Functions

Campbell University

Harvey Mudd College

Baylor University

Lectures

01:43

In mathematics, a function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. An example is the function that relates each real number x to its square x^2. The output of a function f corresponding to an input x is denoted by f(x).

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(a) Draw the graph of the …

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All right, we're looking at this function X plus one squared plus four. And, uh, as long as you know, the parent function wide was X squared is the U shape parabola that starts at the origin. This answer is actually really easy. You should come to your really fast, because when you look at the transformations, you have to remember that it's backwards. You go left, uh, when it's inside the parenthesis, so left one and then up four. And it also has a vertical stretch by a factor of three. But what's important by this problem is that you recognize that when you shift from the origin left one up. Or that's by the way. That's called your Vertex a negative 14 and all it's happening is we're stretching it vertically. So this is what your graph looks like. There's a very lame picture of your graph for letter A because I didn't show you house narrow. This graph is, but you would see like it's going to have a Y intercept at seven, which doesn't look like it, But it does, um, in part B. It appears like there's no X intercepts, and that should make sense because to verify that you have an X intercept, it's a when Y equals zero. If I plug in zero for why and I try to solve this by subtracting four over, cancel that out and divide by three. Well, what I have next is I have X plus one squared equals negative four thirds. Well, when I try to square root a negative, that's impossible because the whether you have a positive here or negative once you square it, you can only get positive answers. So that confirms what the conjecture was, which is that there were no X intercepts. It's impossible to let this function equals zero and get a real solution. Um, so that I guess I should write that out. You get a non real solution, which means that there's no X intercept. You can graphically show that. Yeah,

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