How do you find the vertex and the axis of symmetry of the equation? y = 2x^2 + 28x - 10
2 answers:
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Answer:
See explanation
Step-by-step explanation:
In ΔABC, m∠B = m∠C.
BH is angle B bisector, then by definition of angle bisector
∠CBH ≅ ∠HBK
m∠CBH = m∠HBK = 1/2m∠B
CK is angle C bisector, then by definition of angle bisector
∠BCK ≅ ∠KCH
m∠BCK = m∠KCH = 1/2m∠C
Since m∠B = m∠C, then
m∠CBH = m∠HBK = 1/2m∠B = 1/2m∠C = m∠BCK = m∠KCH (*)
Consider triangles CBH and BCK. In these triangles,
- ∠CBH ≅ ∠BCK (from equality (*));
- ∠HCB ≅ ∠KBC, because m∠B = m∠C;
- BC ≅CB by reflexive property.
So, triangles CBH and BCK are congruent by ASA postulate.
Congruent triangles have congruent corresponding sides, hence
BH ≅ CK.
Answer:
Step-by-step explanation:
From equations (1) & (2), we find:
6x - 8 = - 3x +10
Plug x = 2 in equation (1)
y = 6(2)-8 = 12-8 =4
So,