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

Skilled2021-05-28Added 96 answers

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:$y=\frac{1}{4}{(-2)}^{2}+(-2)-3=-4$

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:

$0=\frac{1}{4}({x}^{2}+4x-12)$

$0=\frac{1}{4}(x-2)(x+6)$

By zero product property,

x=−6,2

So, the xx-intercepts are at (−6,0) and (2,0).

The graph will be:

[Graph]

The y-coordinate of the vertex (using the function) is:

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:

Factor the right side:

By zero product property,

x=−6,2

So, the xx-intercepts are at (−6,0) and (2,0).

The graph will be:

[Graph]