How do you find the vertex and the intercepts for #f(x)= -x^2-3x-6 #?

Question

Eli Assouline · Accepted Answer

#"see explanation"#

#"given the equation of a parabola in standard form "#

#•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0#

#"then the x-coordinate of the vertex is"#

#•color(white)(x)x_(color(red)"vertex")=-b/(2a)#

#f(x)=-x^2-3x-6" is in standard form"#

#"with "a=-1,b=-3,c=-6#

#rArrx_(color(red)"vertex")=-(-3)/(-2)=-3/2#

#"substitute this value into f(x) for y-coordinate"#

#rArry_(color(red)"vertex")=-(-3/2)^2-3(-3/2)-6=-15/4#

#rArrcolor(magenta)"vertex"=(-3/2,-15/4)#

#color(blue)"Intercepts"#

#• " let x = 0, in equation for y-intercept"#

#• " let y = 0, in equation for x-intercepts"#

#x=0toy=-6larrcolor(red)" y-intercept"#

#y=0to-x^2-3x-6=0#

#"checking the value of the "color(blue)"discriminant"#

#Delta=b^2-4ac=(-3)^2-(4xx-1xx-6)=-15#

#"since "Delta<0" then no real solutions"#

#rArrf(x)" does not intersect with the x-axis"# graph{-x^2-3x-6 [-10, 10, -5, 5]}

Hazel Anglin · Answer

undefined To find the vertex and intercepts for the function $ f(x) = -x^2 - 3x - 6 $, follow these steps:

1. **Vertex:** The vertex of a quadratic function in the form $ f(x) = ax^2 + bx + c $ is given by the formula $ x = -\frac{b}{2a} $.

Substitute the coefficients from the given function:
   $$ x = -\frac{-3}{2(-1)} = -\frac{-3}{-2} = \frac{3}{2} $$

To find the corresponding y-coordinate (or the value of $ f(x) $ at this vertex point), substitute $ x = \frac{3}{2} $ into the function:
   $$ f\left(\frac{3}{2}\right) = -\left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) - 6 $$
   $$ f\left(\frac{3}{2}\right) = -\frac{9}{4} - \frac{9}{2} - 6 $$
   $$ f\left(\frac{3}{2}\right) = -\frac{9}{4} - \frac{18}{4} - \frac{24}{4} $$
   $$ f\left(\frac{3}{2}\right) = -\frac{51}{4} $$

So, the vertex of the function $ f(x) $ is $ \left(\frac{3}{2}, -\frac{51}{4}\right) $.

2. **x-intercepts (or roots):** To find the x-intercepts, set $ f(x) = 0 $ and solve for $ x $. These are the points where the graph intersects the x-axis.
   $$ -x^2 - 3x - 6 = 0 $$

Use the quadratic formula to solve for $ x $:
   $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

Substitute the coefficients from the given function:
   $$ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(-1)(-6)}}{2(-1)} $$
   $$ x = \frac{3 \pm \sqrt{9 - 24}}{-2} $$
   $$ x = \frac{3 \pm \sqrt{-15}}{-2} $$

Since the discriminant $ b^2 - 4ac = -15 $ is negative, the function does not intersect the x-axis, so it has no real roots. Alternatively, you could write the solutions as complex roots: $ x = \frac{3 \pm i\sqrt{15}}{-2} $.

3. **y-intercept:** To find the y-intercept, set $ x = 0 $ and solve for $ f(x) $. This is the point where the graph intersects the y-axis.
   $$ f(0) = -0^2 - 3(0) - 6 $$
   $$ f(0) = -6 $$

So, the y-intercept of the function $ f(x) $ is $ (0, -6) $.