f(x)=2x^2+0x-5
Find the vertex
0/4=0=x
Input x into the equation to find y or f(x)
0+0-5=f(x)
(0,-5) = Vertex
Now find a random x coordinate to input into the equation.
Let's do 4.
32+0-5=f(x)=27
It may not look like a parabola due to it being uneven.
Answer:
144
Step-by-step explanation:
f(4)=3*4^2
f(4)=12^2
12*12=144
Step 4 is wrong. the answer is 1/8, she changed sign on 2^(-3) to 2^(3) without moving it to numerator
These are the steps, with their explanations and conclusions:
1) Draw two triangles: ΔRSP and ΔQSP.
2) Since PS is perpendicular to the segment RQ, ∠ RSP and ∠ QSP are equal to 90° (congruent).
3) Since S is the midpoint of the segment RQ, the two segments RS and SQ are congruent.
4) The segment SP is common to both ΔRSP and Δ QSP.
5) You have shown that the two triangles have two pair of equal sides and their angles included also equal, which is the postulate SAS: triangles are congruent if any pair of corresponding sides and their included angles are equal in both triangles.
Then, now you conclude that, since the two triangles are congruent, every pair of corresponding sides are congruent, and so the segments RP and PQ are congruent, which means that the distance from P to R is the same distance from P to Q, i.e. P is equidistant from points R and Q
