How do you write the quadratic function in vertex form given vertex (1,-10) and point (-3,54)?

Question

Hazel Anglin · Accepted Answer

#(y+10)=4(x-1)^2# or #(y+10)^2+1024(x-1)=0#

A vertex form of equation is of the type

#(y-k)=a(x-h)^2# or #(x-h)=a(y-k)^2#, where #(h,k)# is the vertex.

As the vertex is #(1,-10)#, it would be either

#(y+10)=a(x-1)^2# or #(x-1)=a(y+10)^2#

Case 1

If it is #(y+10)=a(x-1)^2#, as it passes through #(-3,54)#, we have

#(54+10)=a(-3-1)^2# i.e. #16a=64# and #a=4# and equation is

#(y+10)=4(x-1)^2#

Case 2

If it is #(x-1)=a(y+10)^2#, as it passes through #(-3,54)#, we have

#(-3-1)=a(54+10)^2# or #-4=4096a# and #a=-1/1024# and equation is

#(x-1)=-1/1024(y+10)^2# or #(y+10)^2+1024(x-1)=0#

graph{((y+10)-4(x-1)^2)((y+10)^2+1024(x-1))=0 [-5, 5, -80, 80]}

Isaiah Anthony · Answer

undefined To write a quadratic function in vertex form given the vertex (1,-10) and a point (-3,54), you can use the formula:

$$ f(x) = a(x - h)^2 + k $$

Where (h, k) represents the vertex. Substitute the vertex coordinates into the equation:

$$ f(x) = a(x - 1)^2 - 10 $$

Then, use the given point (-3,54) to find the value of 'a':

$$ 54 = a(-3 - 1)^2 - 10 $$
$$ 54 = 16a - 10 $$
$$ 64 = 16a $$
$$ a = 4 $$

So, the quadratic function in vertex form is:

$$ f(x) = 4(x - 1)^2 - 10 $$

Isaiah Anthony · Answer

undefined To write the quadratic function in vertex form given the vertex $(1, -10)$ and a point $(-3, 54)$, follow these steps:

1. Use the vertex form of a quadratic function: $ y = a(x - h)^2 + k $, where $(h, k)$ is the vertex.

2. Substitute the vertex coordinates into the equation: $ h = 1 $ and $ k = -10 $.

3. Plug in the given point coordinates $(-3, 54)$ to find the value of $ a $.

4. Once you have $ a $, rewrite the equation with the vertex and the value of $ a $.

By following these steps, you can write the quadratic function in vertex form.