Graph the following quadratic function. First state the vertex, Iine of symmetry, y-intercept, and xintercepts....

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: ${\displaystyle x=-\left(\frac{1}{2\cdot \left(\frac{1}{4}\right)}\right)=-2}$
The y-coordinate of the vertex (using the function) is: ${\displaystyle y=\frac{1}{4}{(-2)}^2+(-2)-3=-4}$
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:
${\displaystyle 0=\frac{1}{4}{x}^2+x-3}$
Factor the right side:
${\displaystyle 0=\frac{1}{4}(x^2+4x-12)}$
${\displaystyle 0=\frac{1}{4}(x-2)(x+6)}$
By zero product property,
x=−6,2
So, the xx-intercepts are at (−6,0) and (2,0).
The graph will be:
[Graph]