How do you find the vertex and the intercepts for #f(x)= -6x^2+ 5x + 18#?

Question

Nathan Axel · Accepted Answer

#"vertex "=(5/12,457/24),x=5/12+-sqrt457/12# #"the equation of a parabola in "color(blue)"vertex form"# is. #•color(white)(x)y=a(x-h)^2+k# #"where "(h,k)" are the coordinates of the vertex and a"# #"is a multiplier"# #"to obtain this form "color(blue)"complete the square"# #y=-6(x^2-5/6x-3)# #color(white)(y)=-6(x^2+2(-5/12)x+25/144-25/144-3)# #color(white)(y)=-6(x-5/12)+457/24larrcolor(blue)"in vertex form"# #color(magenta)"vertex "=(5/12,457/24)# #"to obtain the x-intercepts set y = 0"# #-6(x-5/12)^2+457/24=0# #(x-5/12)^2=457/144# #color(blue)"take the square root of both sides"# #x-5/12=+-sqrt(457/144)=+-sqrt457/12# #"add "5/12" to both sides"# #x=5/12+-sqrt457/12larrcolor(red)"exact values"#

Abigail Allen · Answer

undefined To find the vertex of the quadratic function $ f(x) = -6x^2 + 5x + 18 $, use the formula for the x-coordinate of the vertex, which is $ x = \frac{-b}{2a} $. In this equation, $ a = -6 $ and $ b = 5 $. Plugging these values into the formula, we get:

$$ x = \frac{-5}{2(-6)} = \frac{5}{12} $$

To find the y-coordinate of the vertex, substitute $ x = \frac{5}{12} $ into the function:

$$ f\left(\frac{5}{12}\right) = -6\left(\frac{5}{12}\right)^2 + 5\left(\frac{5}{12}\right) + 18 $$

$$ f\left(\frac{5}{12}\right) = -6\left(\frac{25}{144}\right) + \frac{25}{12} + 18 $$

$$ f\left(\frac{5}{12}\right) = -\frac{25}{24} + \frac{25}{12} + 18 $$

$$ f\left(\frac{5}{12}\right) = -\frac{25}{24} + \frac{50}{24} + 18 $$

$$ f\left(\frac{5}{12}\right) = \frac{25}{24} + 18 $$

$$ f\left(\frac{5}{12}\right) = \frac{25}{24} + \frac{432}{24} $$

$$ f\left(\frac{5}{12}\right) = \frac{457}{24} $$

So, the vertex of the function is $ \left(\frac{5}{12}, \frac{457}{24}\right) $.

To find the x-intercepts, set $ f(x) = 0 $ and solve for $ x $:

$$ -6x^2 + 5x + 18 = 0 $$

Using the quadratic formula:

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

$$ x = \frac{-5 \pm \sqrt{(5)^2 - 4(-6)(18)}}{2(-6)} $$

$$ x = \frac{-5 \pm \sqrt{25 + 432}}{-12} $$

$$ x = \frac{-5 \pm \sqrt{457}}{-12} $$

So, the x-intercepts are $ x = \frac{-5 + \sqrt{457}}{-12} $ and $ x = \frac{-5 - \sqrt{457}}{-12} $.

To find the y-intercept, substitute $ x = 0 $ into the function:

$$ f(0) = -6(0)^2 + 5(0) + 18 $$

$$ f(0) = 18 $$

So, the y-intercept is $ (0, 18) $.