How do you find a quadratic function whose vertex is at the point (2,9) and has the given x intercepts (0.5,0) & (3.5,0)?

Question

Kennedy Adams · Accepted Answer

Express in vertex form as #f(x) = a(x-2)^2+9#
and solve for #a# to find #a = -4#, so

#f(x) = -4(x-2)^2+9 = -4x^2+16x-7# The function has the following form in vertex form: #f(x) = a(x-2)^2+9# for some constant #a# #0 = f(0.5) = a(0.5-2)^2+9 = 2.25a+9# So #2.25a = -9# and #a = -9/2.25 = -4# So #f(x) = -4(x-2)^2+9# #=-4(x^2-4x+4)+9# #=-4x^2+16x-16+9# #=-4x^2+16x-7#

Julia Adams · Answer

undefined To find a quadratic function with the given vertex and x-intercepts:

1. Use the vertex form of a quadratic function: $ y = a(x - h)^2 + k $, where $ (h, k) $ is the vertex.
2. Substitute the given vertex coordinates into the equation: $ h = 2 $ and $ k = 9 $.
3. Use the x-intercepts to find the value of $ a $.
4. Plug in the values of $ h $, $ k $, and $ a $ into the vertex form equation.

Given vertex $ (2, 9) $ and x-intercepts $ (0.5, 0) $ and $ (3.5, 0) $:

1. Using the vertex form, we have: $ y = a(x - 2)^2 + 9 $.
2. To find $ a $, substitute one of the x-intercepts into the equation and solve for $ a $. Let's use $ (0.5, 0) $:
   $$ 0 = a(0.5 - 2)^2 + 9 $$
3. Solve for $ a $:
   $$ -9 = a(0.5 - 2)^2 $$
   $$ -9 = a(-1.5)^2 $$
   $$ -9 = 2.25a $$
   $$ a = -\frac{9}{2.25} $$
   $$ a = -4 $$
4. Substitute $ a $ into the vertex form equation:
   $$ y = -4(x - 2)^2 + 9 $$

So, the quadratic function with the vertex at $ (2, 9) $ and x-intercepts $ (0.5, 0) $ and $ (3.5, 0) $ is $ y = -4(x - 2)^2 + 9 $.

How do you find a quadratic function whose vertex is at the point (2,9) and has the given x intercepts (0.5,0) &amp; (3.5,0)?

How do you find a quadratic function whose vertex is at the point (2,9) and has the given x intercepts (0.5,0) & (3.5,0)?