How do you write a quadratic equation with Vertex at (1,-1) with y-intercept 3?

Question

Hazel Anglin · Accepted Answer

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

Vertex form is #y=a(x - h)^2 + k#, where #(h, k)# is the vertex. From that information, our equation should look like this:

#y=(x-1)^2-1# That's part of the solution, but we still need to include the #y#-intercept somehow. We've filled in all the variables, except for #a#. Let's see what that does on the parent graph, #y=2(x)^2# vs #y=x^2# graph{y=2x^2}

graph{y=x^2}

It looks like the vertex didn't change, but the graph became narrower. And on the first graph, #x=-1, y=2#, but on the second one, when #x=-1, y=1#. So #a# influences the graph, and could be used to find the #y#-intercept

So, if we want our #y#-intercept to be #3#, we should set #a=3#. WAIT!! Our equation #y=(x-1)^2-1# shows that the graph is shifted down #1# unit. So if we want the #y#-intercept to be #3#, we need to move it up one unit. So our equation will really be #y=4(x+1)^2-1#

Just to double check our work, let's graph it graph{y=4(x-1)^2-1}

We were right! Good job

Avery Alexander · Answer

undefined To write a quadratic equation with a vertex at (1, -1) and a y-intercept of 3, you can use the vertex form of a quadratic equation:

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

where (h, k) represents the vertex of the parabola. Substituting the given values, we have:

$$ y = a(x - 1)^2 - 1 $$

To find the value of 'a', we can use the y-intercept. Since the y-intercept is when x = 0, we can substitute this into the equation and solve for 'a':

$$ 3 = a(0 - 1)^2 - 1 $$

$$ 3 = a(1)^2 - 1 $$

$$ 3 = a - 1 $$

$$ a = 4 $$

So, the quadratic equation is:

$$ y = 4(x - 1)^2 - 1 $$