For the equation y<or equal to -x^2+3x+10 I got the vertex (3/2, 67/4). Is this right?

-(x^2 - 3/2) + 10

-(x^2 - 3/2 + 9/4) + 10 + 9/4
-(x - 3/2)^2 + 49/4

Looks like (3/2,49/4) to me

the x is correct, but

y = -(3/2)^2 + 3(3/2) + 10
= -9/4 + 9/2 + 10
= 49/4

remember in -x^2, you have to square, then multiply by -1, so the result is negative

you are confusing -x^2 with (-x)^2

wow - sorry about my typos. Only the 3rd line of my post is correct.

-(x^2 - 3x + (3/2)^2) is better

To determine if the vertex coordinates you found are correct for the equation y ≤ -x^2 + 3x + 10, we need to follow these steps:

Step 1: Rewrite the equation in vertex form.
The given equation, y ≤ -x^2 + 3x + 10, is in standard form. To find the vertex, we need to convert it to vertex form. The vertex form of a quadratic equation is y = a(x-h)^2 + k, where (h, k) represents the coordinates of the vertex.

To rewrite the equation in vertex form, we need to complete the square. Let's start by isolating y:

y ≤ -x^2 + 3x + 10

Step 2: Complete the square for x.
To complete the square, take half of the coefficient of x (which is 3/2), square it (which is 9/4), and add and subtract it within the equation:

y ≤ -(x^2 - 3x + 9/4 - 9/4) + 10

Step 3: Simplify the squared term.
Now, simplify the expression within the parentheses:

y ≤ -((x - 3/2)^2 - 9/4) + 10

Step 4: Distribute the negative sign.
Distribute the negative sign to the terms inside the parentheses:

y ≤ -((x - 3/2)^2) + 9/4 + 10

Step 5: Combine the constant terms.
Combine the constant terms 9/4 and 10:

y ≤ -((x - 3/2)^2) + 49/4

Step 6: Determine the vertex coordinates.
From the equation, we can see that the vertex is at the point (h, k) = (3/2, 49/4).

Therefore, the vertex coordinates you found, (3/2, 67/4), are incorrect. The correct vertex coordinates for the equation y ≤ -x^2 + 3x + 10 are (3/2, 49/4).