FizeauV

2021-05-27

Graph the following quadratic function. First state the vertex, Iine of symmetry, y-intercept, and xintercepts. All values should be exact. Your graph must have at least three labelled points, one of which must be the vertex. $f\left(x\right)=\frac{1}{4}{x}^{2}+x-3$

Bertha Stark

The xx-coordinate of the vertex of y=ax^2+bx+c. From the given, we have a=14, b=1, and c=−3 so: $x=-\left(\frac{1}{2\cdot \left(\frac{1}{4}\right)}\right)=-2$
The y-coordinate of the vertex (using the function) is: $y=\frac{1}{4}{\left(-2\right)}^{2}+\left(-2\right)-3=-4$
So, the vertex is at (−2,−4).
The line of symmetry of y=ax2+bx+c is x=−b2a (same as the xx-coordinate of the vertex and is a vertical line) so:
x=−2
To find the y-intercept, set x=0 and solve for y using the given function:
y=14(0)2+0−3=−3
So, the y-intercept is at (0,−3).
To find the x-intercepts, set y=0 and solve for xx using the given function:
$0=\frac{1}{4}{x}^{2}+x-3$
Factor the right side:
$0=\frac{1}{4}\left({x}^{2}+4x-12\right)$
$0=\frac{1}{4}\left(x-2\right)\left(x+6\right)$
By zero product property,
x=−6,2
So, the xx-intercepts are at (−6,0) and (2,0).
The graph will be:
[Graph]

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